in-exercises-25-28-determine-if-the-specified-linear-transformation-is-a-one-to-one-and-b-onto-jus-2

Question

In Exercises $25-28$ , determine if the specified linear transformation is (a) one-to-one and (b) onto. Justify each answer. 
The transformation in Exercise 2

Ashley Boni · Accepted Answer

So it&#x27;s the matrix that we got from exercise, too. And now we want to see if the mapping ISS want one are onto or both. Also for party wanted one. We know that the, uh, mapping is wonder one if all the columns of a r linearly independent. But since we have a two by three matrix, we have more columns than we do, rose. So the columns definitely have to be linearly independent. Um, so we do have a wonder one napping to check. If it is on to, we need to see if the columns of a span, um are too. So ah, three and four. I believe in the book tells us that the columns of a span are too. If a has a pivot position in every bro. So we&#x27;re going to start reducing this matrix, so ah, we want there to be a zero here, so I&#x27;m gonna take three times the first row and subtract the second row. So our first row stays the same. So one times three is three minus 30 Four times three is 12 minus a negative. Seven is 19 negative. 19 uh, negative. Three times five is 15 15 plus four gives us 19. So now we&#x27;re in echelon form so we can keep going, but the pivots aren&#x27;t gonna change. So we have a pivot here, and we&#x27;re gonna have a pivot here. So since there&#x27;s one in each row, the columns span are too. So this tells us that the mapping is on two.

Michael Jacobsen · Answer

for this example, we&#x27;re going to be analyzing the transformation. T, which maps the vector space, are three into the vector space are, too. It goes from our three the domain because we have three entries here, and it goes to our to because the co domain contains just two entries. Next, let&#x27;s rewrite this in a slightly friendlier format where Row one of the SPECTRE is x one minus five x two plus four x three, then the second row of this factor is x two minus six x three. With this format, we can express T as mapping a vector X where X is a column vector containing x one x two x three into eight times X where the standard matrix A of the transformation T can be found as follows. For column one. Pull the coefficients we see, which are 10 for X one. Calm to pull coefficients again. They are negative 51 and coefficients and column three R four and negative six. So this is our standard matrix A for the transformation. T. Let&#x27;s analyze this matrix a little bit further. If we look at the pivots, there&#x27;s a pivot here and here and this tells us that a has a pivot in every row. Well, when we have a pivot in every row, that implies immediately that T maps are three on to are to let&#x27;s now discuss whether or not this transfer teach transformation T is going to be one toe one. The first noticed. Something is interesting going on here. We have something that&#x27;s considered three dimensional mapping into something that&#x27;s considered two dimensional, where the transfer tick mation t is linear. If this happens, we&#x27;re going to have a lot of vectors mapping into the zero vector and so t cannot be Oneto one. Well, that&#x27;s the intuition behind it. Let&#x27;s also express why T cannot be Oneto one. Using appropriate dirhams. Let&#x27;s go back to our Matrix A and analyze the pivots once again, where this time we&#x27;re more concerned about the columns than Rose. We have a pivot here here, and the fourth column is not ah, pivot column so we can state next A does not have a pivot in every column. What this then tells us that the columns of a are linearly dependent and whenever the columns of a matrix, the standard matrix from a transformation are linearly dependent. It follows immediately that t is not 1 to 1. So it&#x27;s recap on what we&#x27;ve shown here for the standard matrix A. We found that there is a pivot in every row. That means it&#x27;s on on to transformation. But we found if we think of columns, there&#x27;s not a pivot in every column. So it is not 1 to 1. So for on to think rose for one toe, one Think columns.

Michael Jacobsen · Answer

in this exercise, we&#x27;re dealing with the transformation T that&#x27;s going to map are four, since we have four variables to deal with into our for us. Well, since we have 123 and four entries on the right, we know that we can write T in the form of tea of X, where X is a vector containing X one through X four as eight times X, where the Matrix A is given to be zero 100 for column one Calm to is zero 110 calm three is 00 11 and the fourth column is 0001 Let&#x27;s write this in a better echelon form where we moved a zero row into the last row. We have 11 00 0110 then 0011 and last place the zero row down below. Now for this particular matrix, let&#x27;s also analyze the pivots. We have calm 12 and three being pivot columns, and so this is a non pivot. Call him when we see a non pivot column such as this. What that implies is that first, the transformation T is not one 21 The reason we can assert this right off the bat is TF X equals a X, implies t is linear and a leaner transformation is one toe. One if and only if all columns air pivot columns. We had a non pivot column here, so t is not one toe one next. What about onto for onto? We&#x27;re not looking at the calm eyes were looking at the Rose instead, and we say that T is on two or a non to mapping if and only if all rose of that standard matrix A contains ah, pivot. But we know, too, that since Azov size four by four, if it has a non pivot column, it must also have a non pivot row, which is row for in this case. So this implies that t is not and on to mapping. Or you could say this in a different way. You could say T does not map are four onto our four

Ashley Boni · Answer

So we want to see if the mapping from Ah example 14 is want Warner onto so to be 1 to 1. Ah, the columns of a can&#x27;t be, uh, well, nearly dependent. So a one and a two are definitely linearly independent, since one is not a multiple of each other, the only way ah, one will be a multiple off another is if they were contained on the same line. So a is one for one since a one and a two are linearly independent. Check If it&#x27;s onto. We need Thio. Ensure that there&#x27;s a pivot and both columns of a So a is the Matrix a one in the first column in a two in the second. So now So let&#x27;s look at ah, the equation X equals zero. If there&#x27;s no free variable, unless then that means there has to be a pivot. So this is the same as a one times x one plus a two x two where a one a two are matrices and x one and x two are constants. So this will give us two equations we can see from the picture that, um, both coordinates of a one a two aren&#x27;t zero. So none of these coordinates are zero. So there&#x27;s I&#x27;m just gonna call it a 11 times x one plus a +21 which would be the first entry of a two x two equal zero and a 12 x one plus a +22 x two equal zero says to give us the system. And there&#x27;ll be no free variables here because ah, we can solve for X one and x two explicitly, since all these constants are not zero. So there&#x27;s a pivot in both columns, so this is definitely onto.

Ashley Boni · Answer

So we have our matrix A and we want to determine if the mapping is 1 to 1. So if we take our matrix A into Matt Lab, which it suggests we use, we type the first row semi colon, second row semicolon, third row, semi colon and fourth row. And then if we take, um r R e f which is the command for a reduced Rochel on form of A and give it a name, I just called her are for I guess, reduced. Um, it gives us an output of, uh, this matrix 1070 01 negative 90 0001 and the last row of all zeros. So, ah, by theorem 11 a mapping is onto if the equation a X equals zero has only the trivial solution. So that means that ex would have to be the zero vector. Ah, But since there is no pivot in the third call, we know that we have a nontrivial solution. So Ah, the solution two our X equals zero is nontrivial. Therefore, um, the mapping. I guess tea is not want one

Ashley Boni · Answer

s so we have the linear transformation with the matrix A given. And we want to figure out if the mapping ah t is onto. So we need to look and check that there is a pivot in every row. So if we take the Matrix A and put it into Matt Lab, which it suggests we use, um we just input a equals and we type all the entries in the first row and then put a semi colon, second row, semi colon and do the same. So we&#x27;ve input all five rows four and then row five and then close our brackets. And then if you use the command, our equals r r E f of a. This tells us to, um er sells my love to take a kick the deuce row astral form and signed that matrix Thio makes your car anyway, So if you do this, you get an output of one 0005 So we have our first pivot 010 zero, minus four. But her poo for a second row third row is 001 00 So we have our third pivot. Ah, fourth row is 000 11 So we have our fourth and the last row is all zeros. So since there&#x27;s no pivot in our last row, this means that tea is not onto, since the columns don&#x27;t span or five.

Problem

In Exercises $25-28$ , determine if the specified…

Problem 26 Medium Difficulty

In Exercises $25-28$ , determine if the specified linear transformation is (a) one-to-one and (b) onto. Justify each answer.
The transformation in Exercise 2

Answer

a. $A=\left[\begin{array}{ccc}{1} & {4} & {-5} \\ {3} & {-7} & {4}\end{array}\right]$
b. Yes

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

Linear Algebra and Its Applications

Catherine R.

Anna Marie V.

Heather Z.

Caleb E.

Absolute Value - Example 1

Absolute Value - Example 2

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a. $A=\left[\begin{array}{ccc}{1} & {4} & {-5} \\ {3} & {-7} & {4}\end{array}\right]$b. Yes

Linear Algebra and Its Applications

Catherine R.

Anna Marie V.

Heather Z.

Caleb E.

Absolute Value - Example 1

Absolute Value - Example 2

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

Catherine R.

Anna Marie V.

Heather Z.

Caleb E.

a. $A=\left[\begin{array}{ccc}{1} & {4} & {-5} \\ {3} & {-7} & {4}\end{array}\right]$
b. Yes

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

Linear Algebra and Its Applications