# Matrix question

• April 26th 2009, 05:17 AM
Thomas154321
Matrix question
$A = \left( {\begin{array}{rr}
{-10} & {-18} \\
9 & 17 \\

\end{array} } \right)$

a) Find Y such that $Y^{-1}AY$ is diagonal.

b) Find C such that $C^3 = A$

For a I get $Y = \left( {\begin{array}{rr}
2 & 1 \\
-1 & -1 \\

\end{array} } \right)$

I can't figure out b. Does it involve the Cayley-Hamilton theorem at all? I've only used that for solving $C=A^3$ not this way round. Thanks!
• April 26th 2009, 05:25 AM
Rapha
Hello buddy

Quote:

Originally Posted by Thomas154321
$A = \left( {\begin{array}{rr}
{-10} & {-18} \\
9 & 17 \\

\end{array} } \right)$

a) Find Y such that $Y^{-1}AY$ is diagonal.

b) Find C such that $C^3 = A$

For a I get $Y = \left( {\begin{array}{rr}
2 & 1 \\
-1 & -1 \\

\end{array} } \right)$

This is correct, in this case it is

$Y^{-1}AY = diag(-1,8)$

b) I don't know, sorry
• April 26th 2009, 05:39 AM
Thomas154321
Thanks that's reassuring. (Happy)

I've also found another question I can't do:

Give an example of a square matrix over $\mathbb{C}$ which is not similar to a diagonal matrix, and explain why this is the case.

No clue here. I know the two matrices are similar if they represent the same linear map with respect to two different bases, but other than that I don't know.
• April 26th 2009, 06:10 AM
ZeroDivisor
Hi, regarding b):

we know that
$
Y^{-1}AY = \begin{bmatrix}-1 & 0 \\ 0 & 8 \end{bmatrix}.
$

And hence $A = YDY^{-1}.$

Define $C = \begin{bmatrix}-1 & 0 \\ 0 & 2 \end{bmatrix}.$

Then $C^3 = D.$

Also,

$
YCY^{-1})^3 = YCY^{-1}YCY^{-1}YCY^{-1} = YC^3Y^{-1} = YDY^{-1} = A,
$

so
$YCY^{-1}$
is the matrix we are looking for.

If I work this out, I get

$
YCY^{-1} = \begin{bmatrix} -4 & -6 \\ 3 & 5 \end{bmatrix}.
$

Best,

ZD
• April 26th 2009, 06:10 AM
mr fantastic
Quote:

Originally Posted by Thomas154321
$A = \left( {\begin{array}{rr}
{-10} & {-18} \\
9 & 17 \\

\end{array} } \right)$

a) Find Y such that $Y^{-1}AY$ is diagonal.

b) Find C such that $C^3 = A$

For a I get $Y = \left( {\begin{array}{rr}
2 & 1 \\
-1 & -1 \\

\end{array} } \right)$

I can't figure out b. Does it involve the Cayley-Hamilton theorem at all? I've only used that for solving $C=A^3$ not this way round. Thanks!

Let M = diag(-1, 8). Then $M = Y^{-1} A Y \Rightarrow A = Y M Y^{-1}$.

Therefore the answer to b) is $C = Y N Y^{-1}$ where N = diag(-1, 2) since $(Y N Y^{-1})^3 = Y N^3 Y^{-1} = Y M Y^{-1}$.