Why if $B=M^{-1}AM$ the columns of $M$ are eigenvectors of $A$?

Question

I know that $A$ is diagonalisable if exists a diagonal matrix $B$ such that $B=M^{-1}AM$. Somewhere I read that the columns of M are eigenvectors of A. Why is that? Also, is $M$ uniquely defined?

Answer 1

Let $X_i$ be an Eigen vector of $A$ corresponding to $\lambda_i$ and let the Eigen vectors be independent of each other.

$$A{X_i} = \lambda_i{X_i}$$

Let $\color{green}{M = [{X_1} \ {X_2} \cdots {X_n}]}$ which contains the columns with the Eigen vectors of $A.$

So, $$AM = [A{X_1} \ A{X_2} \cdots A{X_n}] = [\lambda_1{X_1} \ \lambda_2{X_2} \cdots \lambda_n{X_n}]$$

$$AM = [{X_1} \ {X_2} \cdots {X_n}]\begin{bmatrix}\lambda_1 & 0 & \cdots & 0 \\0 &\lambda_2 &\cdots & 0 \\ \vdots & \vdots& \vdots & \vdots \\ 0 & 0 & \cdots &\lambda_n \end{bmatrix} $$

$$\color{blue}{AM = MB}$$

$B$ is the diagonal matrix. $M$ is invertible as the Eigen vectors are independent and $\color{red}{|M| \not= 0}$.

So,

$$\color{blue}{M^{-1}AM = B}$$

Now $M$ need not be uniquely defined, as corresponding to an Eigen value we have infinitely many Eigen vectors.

So, we generally normalize the vectors by dividing them by their magnitudes to obtain $B$ as the diagonal matrix with the Eigen values of $A$.

Answer 2

According to Wikipedia: In linear algebra, a square matrix A is called diagonalizable (or nondefective) if it is similar to a diagonal matrix, i.e., if there exists an invertible matrix P such that $P^{−1}AP$ is a diagonal matrix.

It happens that when a matrix is diagonalizable, you can form $P$ by finding matrix $A$'s eigenvectors. You can calculate $P^{-1}$ from this then.

The intuition is that eigenvectors are those that when applied with linear operator indicated by $A$, the operator would only serve as a dilation of the vector. A matrix is diagonalizable essentially means that the eigenvectors serve as a complete basis for which space $A$ as a linear operator takes input from and gives output to.

Think about this intuition and write down the equations in matrix form of this statement. It would look like $Ax = Dx$ where D is diagonal. Then go for the proof from here. (There is one last step I didn't write out for you to figure out for fun.)

Answer 3

This is stitching together various definitions with different perspectives on matrix multiplication.

1. If $A$ and $M$ are matrices such that $AM$ is defined, the $j$th column of the product matrix $AM$ is $A$ times the vector in the $j$th column of $M$.

2. If $M$ and $B$ are matrices such that $MB$ is defined and $B$ is diagonal, then the $j$th column of $MB$ is the $j$th column of $M$ scaled by the $j$th entry along the diagonal of $B$.

If those two facts aren't obvious, try it out with a few small examples to see why.

A vector $m$ is called an eigenvector of $A$ with (scalar) eigenvalue $b$ if $Am = bm$. If there is a basis of eigenvectors $m_1, m_2, \dots, m_n$ with eigenvalues $b_1, b_2, \dots, b_n$, then $A m_j = b_j m_j$. In other words, for each $j$, then $j$th column of $AM$ equals the $j$th column of $MB$. That is, $AM = MB$, which, since $M$ is invertible (remember its columns are a basis), means $B = M^{-1}AM$.

No, $M$ is not uniquely determined by $A$.

• We could permute its columns any way we like, and we would still have $M^{-1} A M$ diagonal.

• Even if the eigenvalues which make up the diagonal elements of $B$ are fixed, any column of $M$ can be scaled by a nonzero constant, and we would still have $M^{-1} AM=B$.

• If any of the eigenvalues appears more than once, we can substitute for each of the columns of $M$ corresponding to a repeated eigenvalue, any linear combination of those eigenvectors.

Answer 4

Note that you have $$AM=MB$$ where $B$ is a diagonal matrix.

Columns of $AM$ are simply the result of multiplying the matrix $A$ by columns of $M$

On the other hand columns of $MB$ are the diagonal elements of $B$ multiplied by the columns of $M$

Thus we get $$AV=\lambda V$$ for each column $V$ of $M$ where $\lambda$ is a diagonal element of $B$

For the uniqueness of $M$ the answer is no because you may multiply an eigenvector by a scalar and it is still an eigenvector.