Matrix Transformation problem

**craig** · May 30th 2009, 07:40 AM

Hi, another question from me again

A transformation $T$ : $R^2 \rightarrow R^2$ is represented by the matrix $A = \left(\begin{array}{cc}4&4\\4&-2\end{array}\right)$ .

Find:

a) The two eigen values for A
b) A cartesian equation for each of the two lines passing through the origin which are invariant under $T$ .

a) This was easy enough, -4 and 6.

b) No idea how to start this, anyone got any pointers?

Thanks

**craig** · May 30th 2009, 07:41 AM

Invariant means that it does not change under the transformation doesn't it? I'm guessing that you use eigenvalues and eigenvectors at some point?

**craig** · May 30th 2009, 07:46 AM

Sorry just one more thing.

Could someone please explain what exactly is meant by:

$R^2 \rightarrow R^2$

Not too sure what this means :S

Thanks again

**HallsofIvy** · May 30th 2009, 07:51 AM

Originally Posted by craig

Hi, another question from me again

A transformation $T$ : $R^2 \rightarrow R^2$ is represented by the matrix $A = \left(\begin{array}{cc}4&4\\4&-2\end{array}\right)$ .

Find:

a) The two eigen values for A
b) A cartesian equation for each of the two lines passing through the origin which are invariant under $T$ .

a) This was easy enough, -4 and 6.

b) No idea how to start this, anyone got any pointers?

Thanks

Originally Posted by craig

Invariant means that it does not change under the transformation doesn't it? I'm guessing that you use eigenvalues and eigenvectors at some point?

It means that points on that line are mapped into points on that line. For eigenvectors, v, [tex]Av= \lamba v[/itex] and, interpreting v as position vector that means Av points in the same direction as v: They lie on the same line as v.

Yes, part b is just asking you to find the eigenvectors corresponding to the eigenvalues and then write the equations of line through the origin in the direction of those vectors. And that's easy: The fact that -4 is an eigenvalue means that there exist non-zero x, y satisfying $\begin{bmatrix}4 & 4 \\ 4 & 2\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}4x \\ 4y\end{bmatrix}$ . Multiplying that out and setting components equal immediately gives two equations that both satisfy y= -2x. That is one of the invariant lines.

**craig** · May 31st 2009, 11:27 AM

Thanks for the reply, could you just verify this for me if possible?

Originally Posted by HallsofIvy

Yes, part b is just asking you to find the eigenvectors corresponding to the eigenvalues and then write the equations of line through the origin in the direction of those vectors.

Using -4:

$\left(\begin{array}{cc}4&4\\4&-2\end{array}\right) \left(\begin{array}{cc}x\\y\end{array}\right) = -4\left(\begin{array}{cc}x\\y\end{array}\right)$

$4x + 4y = -4x \rightarrow y = -2x$

And using 6

$\left(\begin{array}{cc}4&4\\4&-2\end{array}\right) \left(\begin{array}{cc}x\\y\end{array}\right) = 6\left(\begin{array}{cc}x\\y\end{array}\right)$

$4x + 4y = 6x \rightarrow y = \frac{1}{2}x$

Would this be correct?

Thanks again for your time

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