# How to find transformation matrix which converts matrix to simple standard form

I have a matrix A$$\left( \begin{array}{ccc} 0 & 1 \\ a^2 & 0\\ \end{array} \right)$$ Using eigen values, I convert it into simple standard form B: $$\left( \begin{array}{ccc} a & 0 \\ 0 & -a\\ \end{array} \right)$$ How can I find the transformation matrix M which converts this to simple standard form. In other words, how to find M such that $MAM^{-1}=B$. I am actually interested in knowing the nature of the transformation. Also in general how does one find it?

## 2 Answers

1. find it's eigenvalues using $|\lambda I -A| = 0$ equation, you must find them $a$ and $-a$.

Hint: $\lambda^2-a^2 = (\lambda-a)(\lambda+a)$

1. find their corresponding eigenvectors $e_1$ and $e_2$, using the criteria of eigen-value-vector $\lambda_1 e_1 = Ae_1$

$$a\left(\! \begin{array}{c} x \\ y \end{array} \!\right)=\left(\! \begin{array}{c} 0 & 1 \\ a^2 & 0 \end{array} \!\right)\left(\! \begin{array}{c} x \\ y \end{array} \!\right)$$

\begin{cases} ax=y \\ ay=a^2x \end{cases}

so we notice that a set of solutions is of the form $\left(\! \begin{array}{c} x \\ ax \end{array} \!\right)$ let's pick the vector $e_1=\left(\! \begin{array}{c} 1 \\ a \end{array} \!\right)$ (Do the same to find $e_2$)

according to my calculations $$e_1=\left(\! \begin{array}{c} 1 \\ a \end{array} \!\right)$$ $$e_2=\left(\! \begin{array}{c} 1 \\ -a \end{array} \!\right)$$

1. construct the matrix $M$ with those eigenvectors

$$M=\left(\! \begin{array}{c} 1 & 1 \\ a & -a \end{array} \!\right)$$

And you are done $$M^{-1}=\left(\! \begin{array}{c} \frac{1}{2} & \frac{1}{2a} \\ \frac{1}{2} & \frac{-1}{2a} \end{array} \!\right)$$

Now :

$$M^{-1}AM=\left(\! \begin{array}{c} \frac{1}{2} & \frac{1}{2a} \\ \frac{1}{2} & \frac{-1}{2a} \end{array} \!\right)\left(\! \begin{array}{c} 0 & 1 \\ a^2 & 0 \end{array} \!\right) \left(\! \begin{array}{c} 1 & 1 \\ a & -a \end{array} \!\right)$$

$$M^{-1}AM=\left(\! \begin{array}{c} a & 0 \\ 0 & -a \end{array} \!\right)$$

• Are you saying M is a linear combination of eigen vectors? Oct 10, 2014 at 4:44
• @Bosnia yes refer to this Oct 10, 2014 at 4:54
• @Bosnia I'm not sure if it is a typo but you should look for $M$ such that $M^{-1}AM = ...$ and not $MAM^{-1}$ Oct 10, 2014 at 5:28
• Why is $M^{-1}AM$ preferable? Both are valid.
– robjohn
Oct 10, 2014 at 12:14
• @robjohn: who mentioned {preferable} and {not valid} ? I guess I said {possible typo} in the meaning of 50% typo 50% not. After all I gave both $M^{-1}$ and $M$ and enlightened the difference and it's up to him to decide what he's looking for after checking the {possible} typo! Oct 11, 2014 at 1:16

If $a \ne 0$, the eigenvalues of $A$ are $\pm a$; this is tacitly given in the problem, by stipulating the diagonalized form of $A$, $B$, is diagonal with diagonal entries $a$, $-a$, but it is also easy to see since the characteristic polynomial of $A$ is

$\det (A - \lambda I) = (-\lambda)^2 - a^2 = \lambda^2 - a^2; \tag{1}$

the roots are clearly $\lambda = \pm a$. Since they are distinct, the corresponding eigenectors are linearly independent; we can in fact almost without effort write them down; indeed, direct calculation reveals that

$A \begin{pmatrix} 1 \\ a \end{pmatrix} = \begin{pmatrix} a \\ a^2 \end{pmatrix} = a \begin{pmatrix} 1 \\ a \end{pmatrix} \tag{2}$

and

$A \begin{pmatrix} 1 \\ -a \end{pmatrix} = \begin{pmatrix} -a \\ a^2 \end{pmatrix} = -a \begin{pmatrix} 1 \\ -a \end{pmatrix}; \tag{3}$

one can also validate the linear independence of $v_a = (1, a)^T$ and $v_{-a} = (1, -a)^T$ directly; this is an easy exercise working from the definitions so I leave it to the reader. Suppose we set up the matrix

$E = \begin{bmatrix} 1 & 1 \\ a & -a \end{bmatrix} = \begin{bmatrix} v_a & v_{-a} \end{bmatrix}; \tag{4}$

i.e., the columns of $E$ are the eigenvectors of $A$. Then it is easy to see that

$AE = \begin{bmatrix} Av_a & Av_{-a} \end{bmatrix} = \begin{bmatrix} a v_a & -a v_{-a} \end{bmatrix}. \tag{5}$

We next observe that, the columns of $E$ being linearly independent, $E$ is nonsingular, so we have $E^{-1}$ with

$E^{-1} \begin{bmatrix} v_a & v_{-a} \end{bmatrix} = \begin{bmatrix} E^{-1} v_a & E^{-1} v_{-a} \end{bmatrix} = I, \tag{6}$

i.e.,

$E^{-1} v_a = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \tag{7}$

and

$E^{-1} v_{-a} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}. \tag{8}$

Thus

$E^{-1}AE = E^{-1} \begin{bmatrix} Av_a & Av_{-a} \end{bmatrix} = \begin{bmatrix} a E^{-1}(v_a) & -a E^{-1}(v_{-a}) \end{bmatrix} = \begin{bmatrix} a & 0 \\ 0 & -a \end{bmatrix}. \tag{9}$

We see the matrix $E$ of eigenvectors of $A$ diagonalizes $A$ via the prescription $A \to E^{-1}AE$; if one really needs the answer to read $MAM^{-1}$, simply set $M = E^{-1}$ so that $M^{-1} = E$ and proceed from there.

This technique obviously generalizes to matrices of any size $n$, as long as there are $n$ linearly independent eigenvectors, which will always be the case if the eigenvalues are distinct. Finally, we note that the case $a = 0$ is tacitly excluded by the question itself, since $A$ then becomes nilpotent and can't be diagonalized.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!