1. Another matrix transformation

Hi.

A transformation $T$ : $R^2 \rightarrow R^2$ is represented by the matrix $A = \left(\begin{array}{cc}4&-5\\6&-9\end{array}\right)$.

There is a line through the origin for which every point is mapped onto itself under $T$.

Find the cartesian equation of this line.

I have already worked out the two eigenvalues for this matrix, 1 and -6.

Forming the equation:

$\left(\begin{array}{cc}4&-5\\6&-9\end{array}\right) \left(\begin{array}{cc}x\\y\end{array}\right) = \lambda \left(\begin{array}{cc}x\\y\end{array}\right)$

Where $\lambda$ are my respective eigenvalues.

The question I have is which on do I use, using 1 I get $5y = 3x$ and using -6 I get $y = 2x$.

$\lambda = 1$ is associated with point invariant line
So 1 is the value I should use, but what exactly does the above expression mean?

2. Originally Posted by craig
Hi.

A transformation $T$ : $R^2 \rightarrow R^2$ is represented by the matrix $A = \left(\begin{array}{cc}4&-5\\6&-9\end{array}\right)$.

There is a line through the origin for which every point is mapped onto itself under $T$.

Find the cartesian equation of this line.

I have already worked out the two eigenvalues for this matrix, 1 and -6.

Forming the equation:

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

Where $\lambda$ are my respective eigenvalues.

The question I have is which on do I use, using 1 I get $5y = 3x$ and using -6 I get $y = 2x$.

So 1 is the value I should use, but what exactly does the above expression mean?

????
How did $A = \left(\begin{array}{cc}4&-5\\6&-9\end{array}\right)$ mysteriously change into $\left(\begin{array}{cc}4&4\\4&-2\end{array}\right)$?
Is one of those a typo?

Okay, the first has eigenvalues 1 and -6 as you say while the second has eigenvalues 6 and -4 so I assume the second was probably accidently copied from a different problem.

Notice an important difference between this and a similar problem that was posted (perhaps by you?) earlier on this board. Before the problem asked for "invariant lines" which meant, of course, lines that are mapped into themselves. Those turn out to be the lines in the "characteristic directions" or in the directions given by the eigenvectors. But saying that a line is "invariant" doesn't mean that points on that line are mapped into themselves. It means points on that line are mapped into other points on that same line.

Here, you are asked for a line on which "every point is mapped into itself". That is, you are asked for "invariant points" which happen to lie on the line corresponding to eigenvalue 1.

For example, the point (1, 2) is on the line y= 2x. If you apply the matrix to that, $\left(\begin{array}{cc}4&-5\\6&-9\end{array}\right)\left(\begin{array}{c}1 \\ 2\end{array}\right)= \left(begin{array}{c}-6 \\ -12\end{array}\right)$, a different point because each component is multiplied by -6, but on the same line. If we take the point (3, 5), on the line 3y= 5x, and do the same, $\left(\begin{array}{cc}4&-5\\6&-9\end{array}\right)\left(\begin {array}{c}3 \\ 5\end{array}\right)= \left(begin{array}{c}3 \\ 5\end{array}\right)$, exactly the same point: sure, it has been "multiplied" by its eigenvalue, 1!

Any eigenvector gives an "invariant line" but only eigenvectors corresponding to eigenvalue 1 give "invariant points".

3. Hi thanks for the reply.

Originally Posted by HallsofIvy
Okay, the first has eigenvalues 1 and -6 as you say while the second has eigenvalues 6 and -4 so I assume the second was probably accidently copied from a different problem.
Yes that was my mistake sorry about that.

Originally Posted by HallsofIvy
For example, the point (1, 2) is on the line y= 2x. If you apply the matrix to that, $\left(\begin{array}{cc}4&-5\\6&-9\end{array}\right) \left(\begin{array}{c}1\\2\end{array}\right)= \left(\begin{array}{c}-6\\-12\end{array}\right)$, a different point because each component is multiplied by -6, but on the same line. If we take the point (3, 5), on the line 3y= 5x, and do the same, $\left(\begin{array}{cc}4&-5\\6&-9\end{array}\right)\left(\begin {array}{c}3 \\ 5\end{array}\right)= \left(\begin{array}{c}3 \\ 5\end{array}\right)$, exactly the same point: sure, it has been "multiplied" by its eigenvalue, 1!

Any eigenvector gives an "invariant line" but only eigenvectors corresponding to eigenvalue 1 give "invariant points".
I get what your saying, that's the best anyone's explained it! So if you want invariant points you must have an eigenvalue of 1?

Also, the example you gave above is wrong :P, it should be:

$

\left(\begin{array}{cc}4&-5\\6&-9\end{array}\right)\left(\begin {array}{c}5 \\3\end{array}\right)= \left(\begin{array}{c}5\\3\end{array}\right)
$

As

$

\left(\begin{array}{cc}4&-5\\6&-9\end{array}\right)\left(\begin {array}{c}3 \\ 5\end{array}\right)= \left(\begin{array}{c}-13 \\ -27\end{array}\right)
$

Thanks again for the help, really appreciate it

4. Oh and you have a couple of Latex errors, instead of:

Originally Posted by HallsofIvy
\left(begin{array}{c}3 \\
It should be:

\left(\begin{array}{c}3 \\