Refer a friend and earn $50 when they subscribe to an annual planRefer Now

Get the answer to your homework problem.

Try Numerade Free for 30 Days

Like

Report

Let $M_{2 \times 2}$ be the vector space of all $2 \times 2$ matrices, and define $T : M_{2 \times 2} \rightarrow M_{2 \times 2}$ by $T(A)=A+A^{T},$ where $A=\left[\begin{array}{ll}{a} & {b} \\ {c} & {d}\end{array}\right]$ a. Show that $T$ is a linear transformation.b. Let $B$ be any element of $M_{2 \times 2}$ such that $B^{T}=B .$ Find an $A$ in $M_{2 \times 2}$ such that $T(A)=B$ .c. Show that the range of $T$ is the set of $B$ in $M_{2 \times 2}$ with the property that $B^{T}=B$ .d. Describe the kernel of $T$

a. $T$ is a linear transformation.b. $A = \frac { 1 } { 2 } B$c. the range of $T$ consists of all matrices $B$ in $M _ { 2,2 }$ such that $B ^ { T } = B$d. the kernel of $T = \left\{ \left[ \begin{array} { c c } { 0 } & { b } \\ { - b } & { 0 } \end{array} \right] / b \in \mathbb { R } \right\}$

Calculus 3

Chapter 4

Vector Spaces

Section 2

Null Spaces, Column Spaces, and Linear Transformations

Vectors

Missouri State University

Campbell University

Harvey Mudd College

Boston College

Lectures

02:56

In mathematics, a vector (…

06:36

01:48

Let $\mathcal{D}=\left\{\m…

02:06

Let $\mathbf{v}_{1}$ and $…

08:22

Define $T : \mathbb{P}_{2}…

01:23

Let $\mathcal{B}=\left\{\m…

03:34

Let $V$ be a real inner pr…

Let $\left\{\mathbf{v}_{1}…

03:08

Define $\quad$ a linear tr…

04:46

Let $T_{1}: \mathbb{R}^{2}…

04:32

02:33

Let $V$ be a vector space …

in this problem, we're going to be showing that the transformation t that maps square matrices of size to buy two into the set of two by two square major sees defined by TV equals a plus and set transpose is linear now. The words we want to show t is linear To do this. The definition Lanier is a very important one in linear algebra, so it should immediately pull up that definition in the book. That definition tells us that the transformation is linear. Then it must solve two equations. Let's look at Equation one. First. We must show that tea at a vector U Plus a Vector V is equal to U plus TV. In our case, U and V are going to be square, Major sees. So let's denote them by A and B and for our purposes, maybe. Let's first look at the ray and side of equation one t at a plus to yet be by the definition of our transformation, T would be equal to any place that you transposed, plus B plus be transposed, so that would be our right hand side of the equation. Let's look at the left hand side now will manipulate the left hand side of this equation. Trying to get to this format here we can get to hear their work will be concluded that indeed, t A plus B is equal to t a plus t a p. Let's start our solution over here. First, we need to define all of our objects in linear algebra. So will make a statement that says, Let A and Be Being in M two by two, which is the domain of the transformation team. Then, looking at the left hand side of the equation, we're ready to begin. Well, look at the addition of the matrices A and B inside the parentheses and right T Yet a plus B equals next. We have two things to look at when we define this transformation. We took a matrix, any matrix whatsoever, output itself. Plus it's transposed in our situation. The matrix that we're going to be using is the sum of A and B will be equal to itself. Plus, it's transposed. So was a basic format. This is going to be the evaluation of Tia. A plus B. Inside will have a plus B here and a plus B here a swell, so that starts it soft. The next thing to do is we're trying to get to this format of a plus a trance, PLO's plus people speed transposed. But as it stands, a plus B is stuck inside the parentheses due to the transpose operator. So now we need to look at properties of transposed in order to get to through this problem. One important property of transpose is is that we have a summer Metris ease and we take the transpose of that song will be equivalent to taking the transpose of eight first, the transpose of be first and then adding the result. In other words, these Prentice's sees can be dropped by writing a transposed plus be transposed. Then we can drop these practices here to say Ta Plus B is now equal to a plus B plus a transposed plus be transposed and at this stage were well on her way to the solution. Since we do have a plus, a transposed as we're expecting, and now we have B plus be transposed as well to work with here and here, which gives us this format. The rest of the problem is really more of a matter of substitution. So it's right as our next step A plus, a transposed associate id together plus B plus be transposed. Associate together is equal to the prior step just by rearrangement and association. Then this whole thing will be equal to t a. So if we verified that t A results in a plus a transposed plus TFB this verifies equation one since two a plus B can now be expressed through our transitive any of the qualities as t a plus t a p. Now we're on to the second step of verifying linearity for this problem in the step. One interesting feature we have is we're not looking at two different vectors from the domain vector space were stead looking for a single vector to work with and a scaler or rial number C. We're going to start our off our problem this time by saying, let see be any real number and lit in this case a matrix A be in the vector space off to buy two major sees and sub two by two. So our job is much like what we did before. We want to look at the right hand side of this equation and see just what that would look like for us it b c Times T A. That would result in the scaler see times the result of a plus a transposed. However, this time we need to indicate parentheses and right see times a plus, a transposed. This work here in the scratch work side of this solution tells us if we can get to this stage that once again, we'll go back to see times today by substitution. So let's start on the left hand side of that equation and right T at See Time's A. This will be equal to the input, plus the transpose of that same input. So insert C times a here and C times eight Here is well, now we have to once again use properties of transposed. This time, one property of transpose says, If we're dealing with a skill or multiple of a matrix and transposing the entire result, don't be the same result as first taking a transposed and then multiplying that entire metric spicy itself in symbols, we would have see times a just copying from the first part of this term, plus C times a transposed. This is still not quite what we're looking for. We're trying to get See Time's April's a chance pose, but we're getting very close at this stage. In fact, the next thing to do is just as an ordinary algebra. We can factor out the scaler, see and say that this is C Times A plus, a transpose now through substitution. A plus eight transposed is t A. So you could say that this is now see times t at a. That verifies part two of this solution. Whoever work is still not done since in linear algebra, it's always very critical to state our all of our conclusions will say the following This work shows that t is linear, since we were able to verify equations one and two from the definition of linearity that finishes part A of this problem. For a party of this problem, we're going to do something a little bit different. We're going to assume that be is equal to be transposed. Where be is in M two by two, which just says B is once again a square matrix of size. To buy two. For this problem, we want to show or maybe a better way to say it is solved. T a equals B for A. In other words, if we know that B is equal to be transposed, meaning that matrix B a symmetric, we want to know if there's some matrix A from the domain such that t a would happen to be equal to be before we get too far. With this problem, let's take a look at what this equation here might mean. If t a is equal to be, then we would have a plus. A transposed equals B itself, since that's our definition off what the transformation T does to any Matrix? A. So we're trying to solve this equation. However, we need to find some matrix a such that a plus it's transpose is now equal to be. This can be a tricky thing to do with the given information. But one piece of information we have not yet used in this problem is that B is equal to be transposed. If we apply this information correctly, we should be able to guess what that matrix a should be. So one thing to note is, if we put the Matrix B here and here we would come up with the equation where the be transposed would just result in the B from the given information of this problem. So it's set a equal to be. Then, of course, we have two copies of a A and it's transposed, so it's put in a factor of 1/2 here as a scaler. Then if that were the case, we would get in place of a 1/2 B plus in place of a transposed 1/2 be transposed and this must be equal to be somehow some way where once again they're trickle be be transposed. Here is itself equal to be. Of course, this is not a solution. This could should be taken only a scratch. But it gives us the thought for how to write out that full solution. So let's begin here. We will set a equal to 1/2 times be so a scaler. Multiple of that matrix B or B satisfies b equals be transposed. We're meant to solve t a equals B. And we are assuming that this is now our solution. But we have to verify that this is in fact, the case. Let's right then t a would be equal to a plus, a transposed, and that's just by the definition of this transformation. T Now we can make our substitution since we said on the onset of our solution is in fact 1/2 of B. So it's right that in next will have 1/2 of the Matrix B plus 1/2 of the matrix be transposed. Once again, we can use a trick like we saw in part a where skill and multiple of a matrix is not affected by the transpose operator. So we can say that this is now equal to 1/2 times be plus 1/2 times be transposed now at that good position where we can use the given information of the problem that be itself is equal to be transposed and a word of warning with problems like these, typically, information like this would never be given in a problem unless it's somehow relevant. So if we came up with a solution that never used the fact that B is equal to be transposed, we should be very suspicious of that solution. Next. In our case, let's make the substitution here to say that this is now equal to 1/2 of B plus 1/2 of B. And of course this is equal to the Matrix B itself. So we have shown that if a is equal to 1/2 be, then fact t a would be equal to be itself. And that concludes the second part of this problem. For this problem, we're going to be verifying the following statement about the range of this transformation. T We want to show that the range of tea is equal to the set of all vectors be in the co domain M two by two such that B is equal to be transposed. Let's carefully walk through the definition of ranged to see just exactly what we're going to be getting into. So the range of tea for a transformation satisfies the following. It's equal to the set of all vectors be in the domain. W Our domain is two by two. Such that t of X equals B for some vector X in the domain V were for our case of V is m two by two as well. I'm going to color code much of this problem since if the domain and the co domain are different, then our solution has to follow the same structure that we're going to lay out. So let's begin. In fact, part be answered half of this question already. So let's say that as part of our solution part be dealt with an equation t yet a equals B and we saw from part B. If b equals be transposed, then t a will be equal to be for some matrix A in two by two were m two by two comes from the domain in this case. So what have we shown? We've shown that we already know that B is equal to be transposed then that make tricks be must be in the range of tea. This feels like pretty good work, and a common mistake to make here is to assume the work is now complete. However, there is typically two parts. When we're doing with set equality such as this, we now have to reverse engineer the process and say, Well, what if our matrix B came from the range of tea? Then can we show that B is equal to be transposed? If we could do that, then equality is verified and we'll know that the range of tea is in fact a set of all of these major sees that are symmetric. In other words, be equals be transposed. So let's begin with the second part of this solution. I will say Let be be in the range of tea. Rain is range T for short. We have to say, Well, what does that mean? Next will be is in the range of tea. Then we will be able to say t of Excell equal be equal to be for some vector X envy Whoever in our case, we're dealing with the matrices. So X plays the role of a well, say next then for some matrix A in m two by two, which would be, in this case, the domain. We know that today will be equal to be. But now we have to say, What does this mean exactly? Remember, calls to now show that B is going to be equal to be transposed, but it could be tricky to get all the way there. Let's use another portion of this problem that we have not used just yet. The definition of the transformation itself If t of a is equal to be. Then it follows that a plus a transposed would be equal to be itself, Since that's the definition of the transformation. T it takes the Matrix a outputs itself. Plus it's transposed. So you might say, Well, this doesn't look particularly helpful. So let's look at what the transposed might look like of that matrix B. So the scratch work board portion of this solution weaken, say, be transposed would be by substitution, taking a plus a transposed and taking the transpose of that result. Well, we already saw in part a that the transposed process will distribute across matrix addition. So it's right out that as a very next step in our scratch work, this will be equal to a transposed, plus a transposed transposed. That's a very interesting statement to make. So we have to wonder. What does it mean to take the transpose of a transposed matrix? Well, we already know that the transpose have made matrix will interchange the columns of rose. So if we do that interchange twice by taking the transposed twice, we should get the original Matrix A. And in fact, by looking at properties of transposed, that is the case so we can say that this is a transposed plus a itself coming from a transposed transposed. But if we look at this carefully, this is just a rearrangement of the original Matrix B itself that shows be transposed is equal to be. Let's state this next in our official solution will say then be transposed will be a plus, a transposed transposed which is a transposed plus a transpose transposed, which is next. Hey, transpose plus a The order is different, but matrix addition is community tive, So we can now say that this is in fact, equal to be keep in mind once again that leaning algebra. We have to be very careful about her conclusions. So stay all of our conclusions. Next. Let's also see what we've just shown. We've shown that if B is in the range of tea, then just so happens that B is equal to be transposed. But that's what it requires to be in the set on the right hand side of this problem. Yes, we can. L say by this work. This shows that be is in the set of vectors be in m two by two such that B equals be transposed, then. This shows that the range of tea isn't practical to that set B in m two by two such that b equals be transposed. Now that we've stayed all the conclusions, the second part is complete. This was a part involving set equality. So whenever you see in linear algebra a statement about quality and steps, keep in mind that typically involves two parts. The last problem. We verified a statement about the range of this transformation. T In this problem, we want to investigate with the colonel that transformation would be equal to. In other words, we want to determine the following What is the kernel of the transformation teeth for this problem? Let's go back to the definition of what the kernel of a transformation T is equal to. It's said to be the set of all vectors. Exit V such that t of X equals zero, where zero is the zero vector from the co domain. That's right out. What that is your vector would be in this case, zero as a vector comes from the set of to Buy to meet Tracy's, so it must be a square matrix with two rows two columns, and it has only zero entries. So that's what this Euro vector would look like in our case. Now we're ready to get started. After analyzing the definition, the underlying equation and the zero vector for that equation, we'll start by saying the following supposed okay is in the colonel of tea immediately. Once we made this statement that a came from the kernel of tea, this equation is on the table. So let's start by working with that equation, then t at a would be equal to is your vector, which implies that a plus a transposed is the zero vector itself. Since that's the definition of teeth transformation teeth. Now we need to determine all maitresse sees a in that kernel of T. One way to do this is to specify what the entries of that matrix A should look like. So it's insert some additional notation for this specific problem. Let a be equal to a B C D. Going rogue wise now that we have a form for the two by two Matrix A. We could go back to this equation, start manipulating each one of the entries. Let's write out first a plus a transposed will say next, then a, which is a B CD plus a transposed recall that we said in the prior steps that when we take the transpose of a what we're really doing is inter changing rose with columns. So if I go to Row One, which goes a B that will be interchange into a column, so call him one of the transposed matrix is a B. So, in other words, this just gets flipped to a column rather than a row. Similarly, CD gets transposed into CD as a column at the same time this is equal to the zero matrix itself. Let's write it down on additional step and say, If we add together all of these and these two matrices on the left, going entry wise then in the 11 position will have to A in the 22 positions will have to D. Then we'll have a B plus C in the one to position and in the row to call in one position. The 21 position will have C plus B, and all of this is equal to zero matrix containing Onley zero entries. The next step we can analyze is we're dealing with not your ordinary quality, but with equality of maitresse. Ease recall that matrices air equal if, and only if each of the corresponding entries are equal. So this last equation gives us a tremendous amount of information. It tells us to A is equal to zero. We go to the next entry. It tells us also that B Plus C is equal to zero going on down the line. C plus B is also equal to zero and finally, to d corresponding to this entry tells us to D is also zero notice that we're writing down zeros in two different ways. This would be the ordinary zero, and when we write zero this way were indicating an actual vector which came from the CO domain M two by two. Let's come back to these equations from the 1st 2 equations on the left. Here we can conclude her right away that a must be equal to D, which must be equal to zero since two is non zero and the product of two numbers equaling zero means one of the other had to have been zero coming over to these two sets of equations. We can say immediately that be should be equal to negative C. This tells us now that that matrix A, which we identified earlier on in the solution here, can be expressed as follows eighties effect equal to a which is equal to zero in this position de which is equal to zero in this position. So the main diagonal this matrix contains only zero elements. We have a bee in the one to position. Let's copy that in and we have a C in the one. Excuse me to one position. We can say that this C could be re expressed as C equals negative B. So it's insert negative. Be here now that's our matrix. A but linear algebra. Once again, if we leave our answer in this form, we may not get full credit in many classes. We have to fully state our conclusion to the problem on our conclusions should be what is the kernel of tea? So it's right that out. I will say this work shows the kernel of tea is equal to the set of Mitrice's that satisfy the following. They would have been the say of matrix sees X and B from the definition recall the in this case is the domain of matrices. And two by two that's the major sees a we were working with all along and so we can say that matrix a satisfies entries zero b in row one negative B zero NRO to and after the colon. We just had to give a condition on B. The only condition is that be, is any rial number and that finally concludes our problem. We have determined that the kernel of tea is all matrices A in this form.

View More Answers From This Book

Find Another Textbook

In mathematics, a vector (from the Latin word "vehere" meaning &qu…

In mathematics, a vector (from the Latin "mover") is a geometric o…

Let $\mathcal{D}=\left\{\mathbf{d}_{1}, \mathbf{d}_{2}\right\}$ and $\mathca…

Let $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$ be a basis for the vector space $V…

Define $T : \mathbb{P}_{2} \rightarrow \mathbb{R}^{2}$ by $T(\mathbf{p})=\le…

Let $\mathcal{B}=\left\{\mathbf{b}_{1}, \mathbf{b}_{2}, \mathbf{b}_{3}\right…

Let $V$ be a real inner product space, and let $\mathbf{u}_{1}$ and $\mathbf…

Let $\left\{\mathbf{v}_{1}, \mathbf{v}_{2}\right\}$ be a basis for the vecto…

Define $\quad$ a linear transformation $\quad T : \mathbb{P}_{2} \rightarrow…

Let $T_{1}: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ and $T_{2}: \mathbb{R…

Let $V$ be a real inner product space, and let u be a fixed (nonzero) vector…

Let $V$ be a vector space with basis $\left\{\mathbf{v}_{1}, \mathbf{v}_{2},…

01:18

In Exercises $5-8,$ find the coordinate vector $[\mathbf{x}]_{\mathcal{B}}$ …

06:14

Diagonalize the matrices in Exercises $7-20,$ if possible. The eigenvalues f…

03:41

Let $A$ and $B$ be $3 \times 3$ matrices, with det $A=-3$ and $\operatorname…

01:12

Use Exercise 27 to complete the proof of Theorem 1 for the case when $A$ is …

01:39

[M] Construct a random $4 \times 4$ matrix $A$ with integer entries between …

02:00

Find the dimension of the subspace of all vectors in $\mathbb{R}^{3}$ whose …

07:47

[M] Compare two methods for finding the steady-state vector $\mathbf{q}$ of …

03:51

Find the determinants in Exercises 5–10 by row reduction to echelon form.

08:28

Find the particular solution of $(15)$ that satisfies the boundary condition…

92% of Numerade students report better grades.

Try Numerade Free for 30 Days. You can cancel at any time.

Annual

0.00/mo 0.00/mo

Billed annually at 0.00/yr after free trial

Monthly

0.00/mo

Billed monthly at 0.00/mo after free trial

Earn better grades with our study tools:

Textbooks

Video lessons matched directly to the problems in your textbooks.

Ask a Question

Can't find a question? Ask our 30,000+ educators for help.

Courses

Watch full-length courses, covering key principles and concepts.

AI Tutor

Receive weekly guidance from the world’s first A.I. Tutor, Ace.

30 day free trial, then pay 0.00/month

30 day free trial, then pay 0.00/year

You can cancel anytime

OR PAY WITH

Your subscription has started!

The number 2 is also the smallest & first prime number (since every other even number is divisible by two).

If you write pi (to the first two decimal places of 3.14) backwards, in big, block letters it actually reads "PIE".

Receive weekly guidance from the world's first A.I. Tutor, Ace.

Mount Everest weighs an estimated 357 trillion pounds

Snapshot a problem with the Numerade app, and we'll give you the video solution.

A cheetah can run up to 76 miles per hour, and can go from 0 to 60 miles per hour in less than three seconds.

Back in a jiffy? You'd better be fast! A "jiffy" is an actual length of time, equal to about 1/100th of a second.