Basis of the range of a Linear Transformation

[WK95]

Mod note: fixed an exponent (% --> 5) on the transformation definition.

Homework Statement

A is a (4x5)-matrix over R, and L_A:R^5 --> R^4 is a linear transformation defined by L_a(x)=Ax. Find the basis for the range of L_A.

Homework Equations

The Attempt at a Solution

$A = \begin{bmatrix}1 & 2 & 3 & 4 & 5 \\2 & 3 & 4 & 5 & 6 \\3 & 4 & 5 & 6 & 7 \\4 & 5 & 6 & 7 & 8 \end{bmatrix}$

$A = \begin{bmatrix}1 & 2 & 3 & 4 & 5 & b_{1} \\2 & 3 & 4 & 5 & 6 & b_{2} \\3 & 4 & 5 & 6 & 7 & b_{3} \\4 & 5 & 6 & 7 & 8 & b_{4} \end{bmatrix}$

$A = \begin{bmatrix}1 & 0 & -1 & -2 & -3 & -3b_{1}+2b_{2} \\0 & -1 & -2 & -3 & -4 & -2b_{1}+b_{2} \\0 & 0 & 0 & 0 & 0 & b_{1}-2b_{2}+b_{3} \\0 & 0 & 0 & 0 & 0 & 2b_{1}-3b_{2}+b_{4} \end{bmatrix}$

Where do I go from there?

 

[Mark44]

Mod note: fixed an exponent (% --> 5) on the transformation definition.

Homework Statement

A is a (4x5)-matrix over R, and L_A:R^5 --> R^4 is a linear transformation defined by L_a(x)=Ax. Find the basis for the range of L_A.

Homework Equations

The Attempt at a Solution

$A = \begin{bmatrix}1 & 2 & 3 & 4 & 5 \\2 & 3 & 4 & 5 & 6 \\3 & 4 & 5 & 6 & 7 \\4 & 5 & 6 & 7 & 8 \end{bmatrix}$

$A = \begin{bmatrix}1 & 2 & 3 & 4 & 5 & b_{1} \\2 & 3 & 4 & 5 & 6 & b_{2} \\3 & 4 & 5 & 6 & 7 & b_{3} \\4 & 5 & 6 & 7 & 8 & b_{4} \end{bmatrix}$

Let's put some words with the above so that you can understand what's going on. For some b $\in$ R⁴, the augmented matrix above represents this matrix equation:
Ax = b

$A = \begin{bmatrix}1 & 0 & -1 & -2 & -3 & -3b_{1}+2b_{2} \\0 & -1 & -2 & -3 & -4 & -2b_{1}+b_{2} \\0 & 0 & 0 & 0 & 0 & b_{1}-2b_{2}+b_{3} \\0 & 0 & 0 & 0 & 0 & 2b_{1}-3b_{2}+b_{4} \end{bmatrix}$

A bit of row reduction yields the above. Note that I didn't check your work, so it's possible there were some arithmetic errors along the way. Also, you should multiply row 2 by -1 so that the matrix is in proper reduced form.

Where do I go from there?

The first two rows in the matrix tell us about x₁ and the other coordinates of the input vector x. The first row tells us that x₁ is a linear combination of x₃, x₄, and x₅, plus a combination of some of the coordinates of b. The second row tells us that x₂ is a different linear combination of x₃, x₄, and x₅, plus some of the coordinates of b.

The last two rows tell us about the restrictions on b itself, namely that b₁ - 3b₃ + 2b₄ has to be zero, as does b₂ - 2b₃ + b₄. Do you see why?

From these restrictions you can solve for the coordinates of b, and thereby get a basis for the range of your transformation. I have glossed over things a bit here, but if what I've said isn't clear, ask for clarification.

 

[WK95]

Yes, I see why b1 - 3b3 + 2b4 and b2 - 2b3 + b4 have to equal zero. Since they are limiting what sort of values b1,b2, b3, and b4 can take, I'm guessing they have something to do with range?

 

[Mark44]

Yes, I see why b1 - 3b3 + 2b4 and b2 - 2b3 + b4 have to equal zero. Since they are limiting what sort of values b1,b2, b3, and b4 can take, I'm guessing they have something to do with range?

In fact, they define the range.

 

[WK95]

Ah Ok. So the answer to the problem is
Rg(A) = {(b1,b2,b3): b1 - 3b3 + 2b4 = 0, b2 - 2b3 + b4=0}

Is the notation for the solution correct?

Would the basis in this case be two vectors (because there are two pivot columns?) that both fit the above rule for the range? For b1 - 3b3 + 2b4 = 0, b2 - 2b3 + b4=0, the free variables would be b3 and b4.

If i choose b3 and b4 to equal 1 and 2 respectively, b1 and b2 are -1 and 0 respectively.
If i choose b3 and b4 to equal 2 and 1 respectively, b1 and b2 are 4 and 3 respectively.

the two vectors I get are (-1,0,1,2) and (4,3,2,1) assuming I did this correctly. Would these be the basis?

 

[Mark44]

Ah Ok. So the answer to the problem is
Rg(A) = {(b1,b2,b3): b1 - 3b3 + 2b4 = 0, b2 - 2b3 + b4=0}

Is the notation for the solution correct?

Not quite. Each vector b in the range has four coordinates, not three as you show. And rather than showing the two equations, I would give the range as a set of two vectors in R⁴.

Would the basis in this case be two vectors (because there are two pivot columns?) that both fit the above rule for the range? For b1 - 3b3 + 2b4 = 0, b2 - 2b3 + b4=0, the free variables would be b3 and b4.

If i choose b3 and b4 to equal 1 and 2 respectively, b1 and b2 are -1 and 0 respectively.
If i choose b3 and b4 to equal 2 and 1 respectively, b1 and b2 are 4 and 3 respectively.

What is usually done is to choose b₃ = 1 and b₄ = 0 to get one vector, and then choose b₃ = 0 and b₄ = 1 to get the other vector.

the two vectors I get are (-1,0,1,2) and (4,3,2,1) assuming I did this correctly. Would these be the basis?

If they satisfy the two equations you started with, they're fine. My choices would be <3, 2, 1, 0> and <-2, -1, 0, 1>, though. Your two vectors and my two vectors are linearly independent, since each vector in a pair is not a scalar multiple of the other one in the pair. If it turned out that there were three or more vectors in a basis, my technique would be better, since it would be obvious that the three (or more) vectors were linearly independent. This is because in certain coordinate positions, all but one vector would have 0's in that position, so it wouldn't be possible for any vector to be a linear combination of the others.

 

[WK95]

So how about this?
$Rg(A)= \begin{bmatrix}1 \\0 \\-3 \\2 \end{bmatrix}, \begin{bmatrix}0 \\1 \\-2 \\1 \end{bmatrix}$

 

[Mark44]

So how about this?
$Rg(A)= \begin{bmatrix}1 \\0 \\-3 \\2 \end{bmatrix}, \begin{bmatrix}0 \\1 \\-2 \\1 \end{bmatrix}$

Nope. These are just the two rows of your reduced matrix. If you check, you'll see that they don't satisfy the two equations in the b coordinates.

To get your basis vectors, note that the first row stands for b₁ - 3b₃ + 2b₄ = 0, and the second row stands for b₂ - 2b₃ + b₄ = 0,

Solve the first equation for b₁ and the second equation for b₂. b₃ and b₄ are free variables. If you write the four equations, you'll see that <b₁, b₂, b₃, b₄> can be written as a linear combination of two vectors.

 

[WK95]

So
b1 = 3b3 - 2b4 = 0,
b2 = 2b3 - b4=0
b3=free
b4=free

So

$b_{3}\begin{bmatrix}3 \\2 \\1 \\0 \end{bmatrix}+b_{4}\begin{bmatrix}-2 \\-1 \\0 \\1 \end{bmatrix}$

or

$\begin{bmatrix}3 \\2 \\1 \\0 \end{bmatrix},\begin{bmatrix}-2 \\-1 \\0 \\1 \end{bmatrix}$

 

[Mark44]

So
b1 = 3b3 - 2b4 = 0,
b2 = 2b3 - b4=0
b3=free
b4=free

So

$b_{3}\begin{bmatrix}3 \\2 \\1 \\0 \end{bmatrix}+b_{4}\begin{bmatrix}-2 \\-1 \\0 \\1 \end{bmatrix}$

or

$\begin{bmatrix}3 \\2 \\1 \\0 \end{bmatrix},\begin{bmatrix}-2 \\-1 \\0 \\1 \end{bmatrix}$

Bingo! That's exactly what I have.

 

[WK95]

And that's the basis?

Thanks for your help!

 

[Mark44]

Those two vectors are a basis for the range of L.

 

[WK95]

Can you also check my answers for these questions?

Find the basis for the row space and column space.

The Basis of the Row Space is the first 2 rows (row vectors?) from the RREF form and the column space is the first two columns (column vectors?) of the original matrix.

 

[Mark44]

Can you also check my answers for these questions?

Find the basis for the row space and column space.

The Basis of the Row Space is the first 2 rows (row vectors?) from the RREF form and the column space is the first two columns (column vectors?) of the original matrix.

Sounds about right.

 

[WK95]

How would I go about finding the basis of the kernel?

 

[Mark44]

How would I go about finding the basis of the kernel?

Solve the matrix equation Ax = 0, where A is the matrix of your transformation. Operationally, you could have an augmented matrix, with the last column all zeroes, but you can omit it since it cannot change. Get the matrix in RREF form and read off the values for x₁, etc., exactly as you did when you found the range. This time, though, the solution will represent the vectors in R⁵ that get mapped to the 0 vector in R⁴.

The important part to remember is that you're finding the solutions to Ax = 0. Since there will be an infinite number of them, you want to present the kernel as a set of vectors that form a basis for it.

 

[WK95]

So x1=x3+2x4+3x5, x2=-2x3-3x4-4x5.

Would the solution be
Basis of kernel = {$\begin{bmatrix}0 \\0 \\1 \\2 \\3 \end{bmatrix},\begin{bmatrix}0 \\0 \\-2 \\-3 \\-4 \end{bmatrix}$}

 

[Mark44]

So x1=x3+2x4+3x5, x2=-2x3-3x4-4x5.

Yes.

Would the solution be
Basis of kernel = {$\begin{bmatrix}0 \\0 \\1 \\2 \\3 \end{bmatrix},\begin{bmatrix}0 \\0 \\-2 \\-3 \\-4 \end{bmatrix}$}

No.

This is the standard trick, so it pays to learn it.
Write the five equations.

x[SUB]1[/SUB] = x[SUB]3[/SUB] + 2x[SUB]4[/SUB] + 3x[SUB]5[/SUB]
x[SUB]2[/SUB] = -2x[SUB]3[/SUB] - 3x[SUB]4[/SUB] - 4x[SUB]5[/SUB]
x[SUB]3[/SUB] = x[SUB]3[/SUB]
x[SUB]4[/SUB] =        x[SUB]4[/SUB]
x[SUB]5[/SUB] =             x[SUB]5[/SUB]

If you look at this just right, you should see that a vector x in the kernel is a linear combination of three vectors.

 

[WK95]

So
Basis of kernel = {$\begin{bmatrix}1 \\-2 \\1 \\0 \\0 \end{bmatrix},\begin{bmatrix}2 \\-3 \\0 \\1 \\0 \end{bmatrix},\begin{bmatrix}3 \\-1 \\0 \\0 \\1 \end{bmatrix}$}

With those values being from the coefficients of x3, x4, and x5 respectively.

 

[Mark44]

The 3rd vector isn't in the kernel. The other two are. You have a mistake in your row reduction or copied a number incorrectly.

 

[WK95]

x[SUB]1[/SUB] = x[SUB]3[/SUB] + 2x[SUB]4[/SUB] + 3x[SUB]5[/SUB]
x[SUB]2[/SUB] = -2x[SUB]3[/SUB] - 3x[SUB]4[/SUB] - 4x[SUB]5[/SUB]
x[SUB]3[/SUB] = x[SUB]3[/SUB]
x[SUB]4[/SUB] =        x[SUB]4[/SUB]
x[SUB]5[/SUB] =             x[SUB]5[/SUB]

Hmm... i doubled checked and the row reduction is correct. I'm not sure where I went wrong.

I got those three vectors from this which looks correct. How did you deduce that the last vector should not be there?

 

[Mark44]

x[SUB]1[/SUB] = x[SUB]3[/SUB] + 2x[SUB]4[/SUB] + 3x[SUB]5[/SUB]
x[SUB]2[/SUB] = -2x[SUB]3[/SUB] - 3x[SUB]4[/SUB] - 4x[SUB]5[/SUB]
x[SUB]3[/SUB] = x[SUB]3[/SUB]
x[SUB]4[/SUB] =        x[SUB]4[/SUB]
x[SUB]5[/SUB] =             x[SUB]5[/SUB]

Hmm... i doubled checked and the row reduction is correct. I'm not sure where I went wrong.

I got those three vectors from this which looks correct. How did you deduce that the last vector should not be there?

From the way I set up the three equations you can read off the three vectors. I checked my 3rd vector and it works, and yours doesn't. If you stare at what I wrote long enoughtm it should come to you.

 

[WK95]

Basis of kernel = {$\begin{bmatrix}1 \\-2 \\1 \\0 \\0 \end{bmatrix},\begin{bmatrix}2 \\-3 \\0 \\1 \\0 \end{bmatrix},\begin{bmatrix}3 \\-4 \\0 \\0 \\1 \end{bmatrix}$}

Wow.. I missed that. the -1 in the 3rd vector should of been a -4. Stupid me...

 

[Mark44]

Basis of kernel = {$\begin{bmatrix}1 \\-2 \\1 \\0 \\0 \end{bmatrix},\begin{bmatrix}2 \\-3 \\0 \\1 \\0 \end{bmatrix},\begin{bmatrix}3 \\-4 \\0 \\0 \\1 \end{bmatrix}$}

Wow.. I missed that. the -1 in the 3rd vector should of been a -4.

Right.

Stupid me...

As a check, and something you should always do, multiply your matrix times each of the vectors in your basis. Since they are vectors in the kernel, the result of each multiplication should be the zero vector.

 

[WK95]

Right.
As a check, and something you should always do, multiply your matrix times each of the vectors in your basis. Since they are vectors in the kernel, the result of each multiplication should be the zero vector.

Thanks. That's very helpful to know!

 

[WK95]

Additionally, how do I "express 0 as an element of R^4 as a linear combination of the column vectors of A in which every scalar is nonzero?"

For one, I know the kernel is x such that Ax=0. Additionally, I know the basis of the kernel so i know that a certain linear combination of the vectors in the basis of the kernel will get me an answer or so I think. For sure, I'm certain that the kernel is involved somehow.

 

[Mark44]

Additionally, how do I "express 0 as an element of R^4 as a linear combination of the column vectors of A in which every scalar is nonzero?"

For one, I know the kernel is x such that Ax=0. Additionally, I know the basis of the kernel so i know that a certain linear combination of the vectors in the basis of the kernel will get me an answer or so I think. For sure, I'm certain that the kernel is involved somehow.

For your question above, it seems to me that you are asking how to show that the columns of the matrix are linearly dependent. Since there are five columns, and the columns are vectors in R⁴, they have to be linearly dependent. This means that the equation c₁L₁ + c₂L₂ + c₃L₃ + c₄L₄ + c₅L₅ = 0 has a solution in which at least one of the constants, c_i is nonzero. (Here L_i is the i-th column of the matrix.) That's how I understand what you're asking in the first question above.

I don't see much of a direct connection between this and what you say about the kernel. If you had a square matrix (same number of columns as rows) and if the columns were linearly independent, then the transformation that the matrix represents would be one-to-one, and the kernel would necessarily consist of only the zero vector.