Find a matrix ##C## such that ##C^{-1} A C## is a diagonal matrix

[Hall]

Homework Statement

Let A be a square matrix.

Relevant Equations

$C^{-1} AC$

I’m really unable to solve those questions which ask to find a nonsingular $C$ such that
$$
C^{-1} A C$$
is a digonal matrix. Some people solve it by finding the eigenvalues and then using it to form a diagonal matrix and setting it equal to $$C^{-1} A C$$. Can you please tell me from scratch the logic behind the solving those problems?

 

[PeroK]

There must be lots of resources online that show why constructing $C$ from the eigenvectors of $A$ works.

 

[vela]

There must be lots of resources online that show why constructing $C$ from the eigenvectors of $A$ works.

As well as your textbook.

 

[Eclair_de_XII]

Let's say you constructed a matrix $C$ from a basis of eigenvectors. What is the result of applying $A$ to one of the columns of $C$? What, then, happens when you apply $C^{-1}$ to the result?

That being said, I read this somewhere and have no idea how this was derived. But I do know why it works.

 

[Office_Shredder]

Three things to remember here is
- a diagonal matrix is one that maps $e_i$ to a multiple of $e_i$.

- If $v_i$ is the $i$th column of $C$, then $Ce_i=v_i$.

- Also by definition of the inverse, $C^{-1} v_i = e_i$.

Now try applying all the matrices to $e_i$ and see what you get at the end.

 

[WWGD]

@Hall: Notice such matrix C is not guaranteed to exist for all matrices A. Some assumptions must be made on A. Do you know what they are?

 

[Hall]

@Office_Shredder @WWGD Thanks for coming in. A few honorable men developed an impression that I was asking for spoon-feeding, so, it has become imperative now to unbosom what I already know and what I wanted to know:

Let $A$ be a $2 \times 2$ matrix and represents the linear transformation $T : V \to V$, w.r.t. basis $(e_1, e_2)$.

$T$ might have two eigenvalues $\lambda _1$ and $\lambda _2$. Then, w.r.t. eigenvectors $u_1$ and $u_2$, the diagonal matrix
$$
\Lambda =
\begin{bmatrix}
\lambda_1 & 0\\
0 & \lambda_2\\
\end{bmatrix}$$
Will also represent $T$. Thus, $\Lambda$ and $A$ are similar, therefore there exits a nonsingular $C$ such that
$$
\Lambda = C^{-1} A C$$

But in Question number 2 of exercise 4.10 of Apostol’s Calculus Vol II, the author just asks to convert A into a diagonal matrix by combining it thus
$$
C^{-1} AC$$
And my doubt is why do we equate it to $\Lambda$ only and not to any other diagonal matrix, say
$$
\begin{bmatrix}
\alpha &0 \\
0 & \beta\\
\end{bmatrix}$$

 

[nuuskur]

$A$ is diagonalisable if it is similar to a diagonal matrix. It is known that $A$ is diagonalisable if and only if $A=PDP^{-1}$, where $D$ is a diagonal matrix whose diagonal elements are the eigenvalues of $A$. For every eigenvalue, pick an eigenvector corresponding to that value - these eigenvectors are used to make $P$. Suffices to find bases for the individual eigenspaces, for instance.

 

[Hall]

Let's say you constructed a matrix $C$ from a basis of eigenvectors. What is the result of applying $A$ to one of the columns of $C$? What, then, happens when you apply $C^{-1}$ to the result?

That being said, I read this somewhere and have no idea how this was derived. But I do know why it works.

Oh! Thanks. I think I can demonstrate that. Considering $A$ to be a 2 x 2 matrix, and letting its eigenvalues to be $\lambda_1$ and $\lambda_2$. Let the basis of eigenvectors for $\lambda_1$ be $(x_1, x_2)$ and the basis of eigenvectors for $\lambda_2$ be $(y_1, y_2)$. Then,
$$
C =
\begin{bmatrix}
x_1 & y_1\\
x_2 & y_2\\
\end{bmatrix}$$
The first column of $AC$ (of course A times C, not air conditioner)
$$
A \times
\begin{bmatrix}
x_1 \\
x_2\\
\end{bmatrix}
=
\begin{bmatrix}
\lambda_1 x_1\\
\lambda _1 x_2\\
\end{bmatrix}$$

And the second column is
$$
A \times
\begin{bmatrix}
y_1\\
y_2\\
\end{bmatrix}
=
\begin{bmatrix}
\lambda_2 y_1\\
\lambda_2 y_2\\
\end{bmatrix}
$$

Thus, we have
$$
A ~C =
\begin{bmatrix}
\lambda_1 x_1 & \lambda_2 y_1\\
\lambda_1 x_2 & \lambda_2 y_2\\
\end{bmatrix}
$$

Now, multiplying that with $C^{-1}$, we have the first column as
$$
C^{-1} \times \lambda_1
\begin{bmatrix}
x_1\\
x_2\\
\end{bmatrix}
=
\lambda _1
\begin{bmatrix}
1\\
0\\
\end{bmatrix}
$$
And the second column would be
$$
C^{-1} \lambda_2
\begin{bmatrix}
y_1\\
y_2\\
\end{bmatrix}
=
\begin{bmatrix}
0\\
\lambda_2 \\
\end{bmatrix}
$$

Thus,
$$
C^{-1} A C =
\begin{bmatrix}
\lambda_1 &0\\
0 & \lambda_2\\
\end{bmatrix}
$$

Hence, $A$ is diagonalized.

Thanks for giving this idea.

 

[PeroK]

And my doubt is why do we equate it to $\Lambda$ only and not to any other diagonal matrix, say
$$
\begin{bmatrix}
\alpha &0 \\
0 & \beta\\
\end{bmatrix}$$

Matrix diagonalisation is effectively unique:

https://math.stackexchange.com/questions/452320/is-matrix-diagonalization-unique