Finding the Kernel for a Transformation matrix

**lo2** · October 27th 2008, 01:39 PM

I have the following Transformation matrix $F=\begin{bmatrix}1&1&1&0 \end{bmatrix}$ then I have to find the Kernel for it.

I do that by solving the equation:

$\begin{bmatrix}1&1&1&0 \end{bmatrix} \cdot \begin{bmatrix}x_1\\x_2\\x_3\\x_4 \end{bmatrix}=0$ .

But when I multiplicy these two matrices I get a 1x1 matrix.

And I know that there ought to be 3 Kernel matrices (from Maple

)?

SO what do I do?

**ThePerfectHacker** · October 27th 2008, 05:10 PM

Originally Posted by lo2

I have the following Transformation matrix $F=\begin{bmatrix}1&1&1&0 \end{bmatrix}$ then I have to find the Kernel for it.

I do that by solving the equation:

$\begin{bmatrix}1&1&1&0 \end{bmatrix} \cdot \begin{bmatrix}x_1\\x_2\\x_3\\x_4 \end{bmatrix}=0$ .

But when I multiplicy these two matrices I get a 1x1 matrix.

And I know that there ought to be 3 Kernel matrices (from Maple

)?

SO what do I do?

This gives you,
$x_1+x_2+x_3=0 \implies x_1 = - x_2 - x_3$ .
Now if you let $x_4 = s,x_3=t,x_2=r$ then $x_1 = - r - t$ .

Then the kernel is $\{ (-r - t, r,t,s)|r,t,s\in \mathbb{R}\}$

**lo2** · October 28th 2008, 07:43 AM

Originally Posted by ThePerfectHacker

This gives you,
$x_1+x_2+x_3=0 \implies x_1 = - x_2 - x_3$ .
Now if you let $x_4 = s,x_3=t,x_2=r$ then $x_1 = - r - t$ .

Then the kernel is $\{ (-r - t, r,t,s)|r,t,s\in \mathbb{R}\}$

But when I use Maple to generate NullSpace I get 3 vectors:

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

How come?

And another question I have to find a Linear Transformation $g: R^{2x2} \rightarrow R$ where U (the amount of 2x2 symmetric matrices where $A=A^T$ ). But How can I do that when kernel is a 1x1?

**ThePerfectHacker** · October 28th 2008, 08:24 AM

Originally Posted by lo2

But when I use Maple to generate NullSpace I get 3 vectors:

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

How come?

That is because Maple is giving you a basis for the nullspace. What we wrote above was correct. We shown that the nullspace is:
$\left\{ \begin{bmatrix} -r-t\\r\\t\\s \end{bmatrix}: r,s,t\in \mathbb{R} \right\}$

Now,
$\begin{bmatrix} -r-t\\r\\t\\s \end{bmatrix} = r\begin{bmatrix} -1\\1\\0\\0\end{bmatrix} + t \begin{bmatrix}-1\\0\\1\\0 \end{bmatrix} + s \begin{bmatrix} 0\\0\\0\\1 \end{bmatrix}$

Thus, we see $\left\{ \begin{bmatrix} -1\\1\\0\\0\end{bmatrix}, \begin{bmatrix}-1\\0\\1\\0 \end{bmatrix}, \begin{bmatrix} 0\\0\\0\\1 \end{bmatrix} \right\}$ is a basis for the nullspace.

**lo2** · October 28th 2008, 09:00 AM

I would say that

kerf = span{(-1, 1, 0, 0), (-1, 0, 1, 0), (0, 0, 0, 1)}

is a satisfying answer. Agree?

And another question I have to find a Linear Transformation $g: R^{2x2} \rightarrow R$ where U (the amount of 2x2 symmetric matrices where $A=A^T$ ). But How can I do that when kernel is a 1x1??

**ThePerfectHacker** · October 28th 2008, 09:13 AM

Originally Posted by lo2

I would say that

kerf = span{(-1, 1, 0, 0), (-1, 0, 1, 0), (0, 0, 0, 1)}

is a satisfying answer. Agree?

Yes

And another question I have to find a Linear Transformation $g: R^{2x2} \rightarrow R$ where U (the amount of 2x2 symmetric matrices where $A=A^T$ ). But How can I do that when kernel is a 1x1??

I am confused here. By $\mathbb{R}^{2\times 2}$ you mean the space of all $2\times 2$ symmetric matrices? If so you can always let $g: \mathbb{R}^{2\times 2}\to \mathbb{R}$ be the trivial mapping i.e. $g(A) = 0$ . Then $g$ is certainly a linear transformation.

**HallsofIvy** · October 28th 2008, 09:42 AM

Originally Posted by lo2

I would say that

kerf = span{(-1, 1, 0, 0), (-1, 0, 1, 0), (0, 0, 0, 1)}

is a satisfying answer. Agree?

And another question I have to find a Linear Transformation $g: R^{2x2} \rightarrow R$ where U (the amount of 2x2 symmetric matrices where $A=A^T$ ). But How can I do that when kernel is a 1x1??

Is there something missing after "where U"? Where U what? And I don't know what you mean by "kernel is a 1x1". The kernel of any Linear Transformation from vector space A to vector space B is a subspace of A.

Are you required to find a transformation that maps 2 by two matrices to a single number and has "all symmetric matrices" as kernel? (You say "2x2 symmertric matrices where $A= A^T$ but that is true of all symmetric matrices.)

**lo2** · October 28th 2008, 12:29 PM

Well I will try and explain it.

I have this linear transformation $g: R^{2x2} \rightarrow R$ . And I have this subspace U (a subspace of $R^{2x2}$ ) that is all 2x2 diagonal matrices where $A=A^T$ .

Then I have to find a linear transformation $g: R^{2x2} \rightarrow R$ where the kernel is U.

But U is a 2x2 matrix and a such transformation gives a 1x1 matrix...?

What to do!?

**lo2** · October 29th 2008, 07:35 AM

Sorry but I really need help soon...!

So I am kind of desperat and I would really like some help!

**HallsofIvy** · October 29th 2008, 10:43 PM

Originally Posted by lo2

Well I will try and explain it.

I have this linear transformation $g: R^{2x2} \rightarrow R$ . And I have this subspace U (a subspace of $R^{2x2}$ ) that is all 2x2 diagonal matrices where $A=A^T$ .

Then I have to find a linear transformation $g: R^{2x2} \rightarrow R$ where the kernel is U.

But U is a 2x2 matrix and a such transformation gives a 1x1 matrix...?

What to do!?

So you want a linear transformation that maps precisely the symmetric matrices to 0. That is, you want a function that maps the matrix $\left[\begin{array}{cc}a & b \\ c & d\end{array}\right]$ to 0 if and only if b= c. What about taking it to b- c? Is that a linear transformation?

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