show-that-a-i_na-when-a-is-an-m-times-n-matrix-hint-use-the-column-definition-of-a-i_n

Question

Show that $A I_{n}=A$ when $A$ is an $m 	imes n$ matrix. [Hint: Use the (column) definition of $A I_{n} . ]$

Lucas Finney · Accepted Answer

okay for question 32. Here we want to show that Yeah. Matrix a times The n identity matrix is equal to a when a is an m upset. Getting ahead of myself is an M by n matrix on so matrix a times the end identity matrix by the definition of matrix multiplication the same thing as a times And now I&#x27;m going to write Oh, actually, yeah. Sorry is going Teoh, right? It the wrong way. We have e one as the basis the first basis of our in dimensional space. So that would be the 10000 etcetera Vector to up Thio n by the definition of matrix multiplication that is the same thing as A and e one. A time to eat too, all the way up to a 10 e n. Now, if we consider one of those columns E one Well, that&#x27;s the same thing is a times 1000 dot, dot dot Which when we do our matrix are multiplication of the Matrix times. This vector we would get that is a 11 a 21 a 31 up to eight times and one We&#x27;re multiplying the first row element or sorry, The first column element of each row by one and every other column element by zero and similarly e N is just going to end up being a n one. A n two dot dot, dot dot hopes. You should write this down to being not a n n So our matrix a tends the identity matrix just going to give us oh, and also actually specify that is the same thing as our A one vector. And that down there is the same thing as R A N Factor and every other one in between will be similar. We have a eight times e one going to be our A one vector. Our first column vector of a eight times that you two is going to be our second column Vector of a and eight times e N is going to be our 10th column vector of a, which by definition, that&#x27;s just Hey, don t o g m matrix bracket. By definition, that&#x27;s just a

Griffin Johnston · Answer

suppose we had on Matrix A. That&#x27;s him by in So it has m rose and in column. So M in columns. Let me just emphasize this year has m number of rows in him Rose and in Cohn&#x27;s in calling. And suppose this major is a was such that the column space of a was equal to the null space of a no what if the null space of a, well, the null space of a is all the set containing all the vectors x all the vectors x such that a times a vector acts is equal to the zero factor. So it&#x27;s the set of all vectors x such that a Times X equals zero vector. So one vector that will always be in the know space of any matrix is that the zero vector the in by 10 factor because the zero vector times A is going to be equal to zero by how matrix mortification works. But another thing to keep in mind here. Another thing to keep in mind is how matrix made more application works. In this particular case, matrix multiplication is only defined if X is this has dimension in by one X must have to mention in by one. It must have the number of the same number of rows as a has columns, since A has in columns X must have in rows. Now, in this video in this video, with all of that in mind, we&#x27;re gonna prove that if the confidence of a is equal to the normal space of a, then this implies this implies that em is equal to in. So this is what we&#x27;re going to prove prove If Colin space of a sequel to the Northwest of a then that implies it in is equal to end. You might already already see why this is true. But let me let me let you know of a nice little trick that or maybe not not even a trick, but a nice, sometimes strategy or initial approach to prove problems like this. If you don&#x27;t see an initial direct path to proving this, if you don&#x27;t he up if it&#x27;s not obvious to you and it&#x27;s OK if I wasn&#x27;t obvious to me as well. But if it&#x27;s not obvious at first sight a direct proof of a direct, direct way of proving this statement then I encourage you to consider the proof by contradiction. So what I mean by that is consider what would happen if M is not equal to end. Suppose M is not equal to end. So in let&#x27;s let&#x27;s try to derive a chain of implications. Let&#x27;s try to derive a chain of implications that leads us to a contradiction. And if we&#x27;re led to a contradiction, If we were led to a contradiction, then that means that this supposition cannot possibly be true. Because if we were to suppose that M is not equal to end, then that would imply a contradictory statement. So am cannot possibly be equal to end. So therefore him must must equal in. So let&#x27;s consider what happens when him is not equal to end. Let&#x27;s first what what subspace does The column space of a lie in If a is a riel has really number entries. A has really number entries. What subspace. There&#x27;s a column space of a lion. Well, the cost base of A is the span of the contractors, and since there are in rows, each column vector has em in trees. So the column space of a is a subspace off our in because each vector in the common space of a is the linear is a linear combination of the column vectors of the Matrix A and each of the column vectors in the Matrix A have em real number entries. So any linear combination of a bunch of vectors that have em real number injuries will also have a real number entry. So each factor in the con space of a we&#x27;ll have em. Leo number entries are therefore each veteran. The confidence of A is in our M the set of all vectors having him Rio number injuries. Now what subspace doesn&#x27;t know space live in the no space, the no space. Well, pause a video and try to think about what subspace been North based lives in. So I&#x27;m assuming you&#x27;ve had a go at it. The no space lives in our in. Why does the North base live in our in? Because the nose basis defined to be the set of all vectors acts such that a times acts is equal to the zero vector. So all the factors in the set X have in rial number in trees, all vectors in this set X have in Rio number interest because the matrix multiplication right here is on Lee defined if X has the same number off entries as there are columns in a so X must have in real number injuries. So check this out. What happens if the column space of a is a self face of RM? Well, this is always gonna be true in the no space of ese subspace are in. This is always gonna be true for an M by N Matrix. But what happens if we were to suppose that him is not equal to end? Well, we we were given the fact that the column space of a is equal to the no space of a. That means that every vector in the column space of A is also in the north face of a and every vector in the north face of A is also in the column space of a But check this out. This can&#x27;t possibly be true. Our statement that we were given can&#x27;t possibly be true if the column space of a lives in our em and the north face of eight lives and are in because the concepts of a lives in a completely separate vector space than the north face of a they live in. You can think of these two vector spaces is completely different universes when in is not equal to em. These are two different universes, so we can see that the column space if they can&#x27;t share any vectors in common with the north face of a if M is not equal to end. But that implies that were contradicting roots groups. That implies that we&#x27;re contradicting this statement right here. So in cannot possibly be not equal to end him must be equal to end, because otherwise we would arrive at this contradiction right here.

Bryan Lynn · Answer

Okay, So here were given to N by N. Matrix sees a matrix, which is a 01 matrix, and the identity matrix, which then we&#x27;ll just have ones on the diagonal and zeros everywhere else. Okay, We want to take the bullion product of thes two matrices. And so what that entails we call the boom part. It is much like a regular product, but where we&#x27;re replacing Addition. Bye. The your operation and multiplication by the end operation. So, for example, in our product matrix, Okay, if we wanted to take the element in the first row, first column will take the forest road, eh? And the first call of I, right? And so first we would support, since we&#x27;re replacing multiplication by the ants. Well, then take a 11 and one. Okay, then our conditions were placed by the or someone we move on to. The next elements would be a one, too. But everything else zeros, right? So it&#x27;s gonna be a 12 and zero, etcetera, etcetera, until we get to a one. And it said at answer anyone and and zero. But oh, these all the rest of it it is just gonna be zeros, right? So the only one that has a potential to be a value other than zero is a 11 and one. Okay. And then we repeat that process, right? So we could take a new row and same column or a same wrote a new column, eyes good mood over one column. Then we would get initially a 11 and zero, but the next value would be a 12 and one, and the rest of Beale zeros, right. So the next value within a one two Sami a 12 and 1 may be considered. See a pattern here, right? Where Everything in the being zeros except for the initial men of a And of course, these whatever a one want everyone to is whatever ay, ay Jay is if it&#x27;s a one when you get one on one so people one. If it&#x27;s a 0011 equals zero, right? So we were really getting. Is that each element? It&#x27;s going to be a I J and one matrix bullion powder today, and I and this is going to be equal to ay, ay jay. Okay, which is also just equal to a

Wendy Wang · Answer

for this question, we have a which is an n by n matrix and were asked to prove that eight times I envy is gonna die And I&#x27;m saying which is simple Let&#x27;s remember what I Ennis I end This is M by m matrix sista. It&#x27;s J entry Look denoted by I i j is equal to zero If I is not called J bicycle to one and I s c culture J to visualize I am is, um, a matrix. So shall we have once on the diagonals and zero everywhere else Something like this. So how do we show a Comes I end Secret I am comes days ago to a Let&#x27;s show it by steps. So let&#x27;s first show eight times i n issue culture A Let&#x27;s to make it more convenient. Let&#x27;s call this from this matrix B s 02 matrices are equivalent if they&#x27;re able to each other if every entry is equal to each other. So if we can show that the i j entry be, it&#x27;s the I J entry. Okay, we are done. So let&#x27;s see. What is the ij a country of B B i j What is it using the definition uses a formula for matrix multiplication. We see that since be usable to a comes I am B I j a sickle. Jule, eh? I won. I won Jae plus a um I chewed. Plus I true j plus a i three. Oh, sorry. This is a typo. This is supposed to be multiplication, right? Here they are, Two times I true, J um, plus a three times i times I three j plus all the way to a high end times i nj So looking at the entries for I m If I saw who deejay at the I I j turn will be zero. So for for every single one of these terms, the size one the I I J term will be dear. Oh, so we&#x27;re left with three only turned us nausea is when eyes equal to J And that is the case. Um, if we have i, Jane Jane because it&#x27;s we&#x27;re always working with the jade calm of I. And what is the corresponding a term? Well, um, if we&#x27;re in the Jake, uh, well, in the Jade Colin, we must have a i j. Now Jay is equal D j so this term becomes one. So essentially, we&#x27;re dealing with a ay, Jay comes one which is just a guy j here. We have shown that B i j a secret you&#x27;re I j on Because the i nj are arbitrary. We showed that every entry off, eh is equal to every entropy, Which is Theo entry off a i m. So from here, we can conclude that because their a i m is equal to a entry by entry A I am isn&#x27;t equal to a Now let&#x27;s try to show I end him. Say, is he called you a That&#x27;s called this matrix. See what ISS The I J entry of C again, we can use the, uh, formula for matrix multiplication. See, I Jay has I I want times a won jae close. Aye, aye. Two times a to J. Plus, I I three times a three j So on and so forth until we get I high end times, eh, Andrzej? Okay. Every single one of the&#x27;s I turn will be zero, except for one of, um so which one of this? Well, it is when a choo index is off. I are the same and that is the case. If we have, I I Because look, we&#x27;re also we&#x27;re always working the I throw off, I I throw off. I m, um, and the course funding. Oh, sorry. These are type of resistible to lower case A. Sorry about that Confusion. Because we&#x27;re talking about the entry. What is the corresponding that, um, country in matrix eight. Well, if we&#x27;re in the eye column, we must have a i J. Because we&#x27;re always is that we&#x27;re always in the Jade column off, eh? So when I as you could you I buy the definition off i n This term will be one. So this will be someone We will have one times a j which is straight just equal to ay, ay, jay. So we&#x27;ve shown that this I j term off matrix C is equal to ay jay Term off matrix eight. Since I nj are arbitrary, every entry of make sexy is to every entry off. Um matrix a Now, um, since cuz you&#x27;re i n times a we&#x27;ve shown that I am hums a is equal to a and she brought entry. And that is how we can conclude I end comes a is equal to a

Michael Jacobsen · Answer

in this video, we&#x27;re considering a matrix A and what we have is that a is written down in this notation. Here were a one through a n or the columns of a There&#x27;s an columns. Africanus Assume that each column has m entries. So that overall Hamza M by N matrix. Now what we&#x27;re going to do next is consider this matrix, which is I am. This is going to be a square matrix of M rose and in columns, which is also the identity matrix. That means we can write it in column form using the same kind of notation with the one you too up through E I m. As this way. So this is our identity matrix and recall if we pick any one of these columns, E one just means we have a matrix for excuse me, a vector with M entries and one will be in entry one for this 11 is going to be in the last entry em one will be in the second entry of this vector and so on. So what we&#x27;re going to be do doing next is multiplying the identity matrix. I am with our metrics, a Let&#x27;s write out what this would look like using our rule for matrix multiplication. You would take the entire matrix I I M and multiply by the first column of a, which is a one. Then the second calm of the product is going to be. I am times the second calm of a, which is a step, too. Then lastly, we go off to the last column of the product I am times A. And that would be the identity matrix times the last column of a a sub in. So this is what the product would look like so far, but also recalled that the identity matrix multiplying any vector always gives us back that same vector. So in our case, when we look at these products here, this is really just overall in a one. Let&#x27;s write that out in our next step. So we have died. Any matrix times A is now equal to a one A to through a sub en. Since thes identity matrices do not change the Matrix Times, the vector product, it just gives us back those vectors. But now that we get to this stage, recall that this is the definition of the Matrix A. So we have just shown that we take an M by N i DNT matrix multiplied by a raise of size M Then we get back just that matrix A itself.

Lucas Finney · Answer

for this problem were given a Matrix A, which is an M by N Matrix in the Matrix D, which is a diagonal matrix with D one to D n along the diagonal. We want to prove that the product a D is found by multiplying the Jade column Vector of a by element J from this D matrix, where one is less than or equal to Jay is less than or equal to end. So first off, I know that because D is a diagonal matrix, the I J element of D is equal to D j times the chronic er Delta i j. So knowing that we can look at the index definition of the matrix product. So the IJ eighth element of a D is going to equal to the sum from K equals one. K equals one up to N of. We want element A i k times element d k j Now, using the definition that we have above for the I j element of d, we can substitute that in. So we have element ai que times d. So we keep J the ending index. But instead of having I you know we have k in the place of I was going to be d j times Delta. Okay, J Now, because of the chronicler Delta here, this is all of the terms of this summer going to be zero except for one. And that is when K is equal to J. So that&#x27;s actually going to be the same thing as element A i j times de J. Now, if we think about if we think about fixing J to a particular value so you know, if we think about the let&#x27;s call it l if we will think about the health column vector of a d. Well, the elements along that elf column vector you&#x27;re going to look like so we would have d l comes a, um yeah, sorry, my blank for a second there d l a times a one l de el times a to l de el times a three l and so on. Or if we think about going from left to right, we would have d one a 11 d one a 21 Do you won a 31 and next with D to a Sorry, that should be one, too. De tu a Tu tu de to a 32 and so on, so we can start to see that pattern of the column vectors in the resulting matrix. The column vectors in a d are just the call in vectors from a multiplied by the de element corresponding to which column were in thus proving thes statement.

Problem

Prove Theorem 3$(\mathrm{d}) .\left[\text { Hint:…

Problem 32 Easy Difficulty

Show that $A I_{n}=A$ when $A$ is an $m \times n$ matrix. [Hint: Use the (column) definition of $A I_{n} . ]$

Answer

$A I_{n}=A$

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David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications

Grace H.

Catherine R.

Anna Marie V.

Heather Z.

Absolute Value - Example 1

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$A I_{n}=A$

Linear Algebra and Its Applications

Grace H.

Catherine R.

Anna Marie V.

Heather Z.

Absolute Value - Example 1

Absolute Value - Example 2

David C. Lay, Steven R. Lay, Judi J. McDonaldLinear Algebra and Its Applications

Grace H.

Catherine R.

Anna Marie V.

Heather Z.

David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications