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For Diagonalizable matrix

I wonder why/how someone came up with the formula:

$$\mathbf{P}^{-1}\mathbf{AP}=\mathbf{D}$$

and not something else like following:

$$\mathbf{AP}=\mathbf{D} $$

$$\text{or}$$

$$\mathbf{PA}=\mathbf{D}$$

Community
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That Guy
  • 1,309

3 Answers3

2

Let ${X_1}, {X_2} \cdots {X_n} $ be independent Eigen vectors of a square matrix A corresponding to the Eigen values $\lambda_1, \lambda_2, \cdots , \lambda_n$.

Let $P = [{X_1} \ {X_2} \cdots {X_n}]$ be the matrix with columns containing Eigen vectors.

Also we have, $$A{X_i} = \lambda_i{X_i}$$

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

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

$$AP = PD$$

where $D$ is the diagonal matrix. $P$ is invertible as the Eigen vectors are independent and $|P| \not= 0$.

So,

$$P^{-1}AP = D$$

We can compute the higher indices of $A$ from the above equation.

$$D^2 = P^{-1}A^2P$$

$$D^3 = P^{-1}A^3P$$

$$\vdots$$

$$D^k = P^{-1}A^kP$$

or

$$A^k = PD^kP^{-1}$$

So, it is conventional to write it as $P^{-1}AP = D$

19aksh
  • 12,768
1

For a linear transformation $A$ on a vector space (finite) $U$, we know that $A$ can be `represented' by a matrix. We usually denote this matrix by $[A]_{\mathcal{B}}$, where $\mathcal{B}$ is a basis for $U$.

Taking the basis for $U$ consisting of eigenvectors of $A$ (when this is possible, i.e. when $A$ has $dim(U)$ linearly independent eigenvectors), gives a particularly simple (diagonal) $[A]_{\mathcal{B}}$ (this is your $D$).

AnyAD
  • 2,594
1

The defining equation for eigenvectors is $A\mathbf v=\lambda\mathbf v$. If we have a bunch of eigenvectors $\mathbf v_i$ with associated eigenvalues $\lambda_i$, using basic properties of matrix multiplication we can collect up all of the individual equations into a single “bulk” equation $$A\begin{bmatrix}\mathbf v_1 & \cdots & \mathbf v_n\end{bmatrix} = \begin{bmatrix}\mathbf v_1 & \mathbf v_2 & \cdots & \mathbf v_n\end{bmatrix} \begin{bmatrix}\lambda_1 & 0 & \cdots & 0 \\ 0&\lambda_2&\cdots&0 \\ \vdots & \vdots & \ddots & \vdots \\ 0&0&\cdots&\lambda_n\end{bmatrix},$$ or $AP=PD$ for short. Note that the matrix of eigenvalues right-multiplies the eigenvector matrix: we want to multiply each column of $P$ by the appropriate value.

If $A$ is diagonalizable, we can choose eigenvectors $\mathbf v_n$ such that they form a basis for the space, in which case $P$ will be square and invertible, so we can multiply both sides by $P^{-1}$ to get $P^{-1}AP=D$. I often use the equation $AP=PD$ as a starting point because I can never remember which side the inverse matrix goes on and it’s easy to derive from the fundamental eigenvector equation.

amd
  • 53,693
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    An easy way to remember is by noting that $D $ operates on coordinate vectors with respect to basis of eigenvectors. So on the LHS of the matrix equation we need a vector expressed as coordinate vector wrt basis consisting of eigenvectirs The columns of this transition matrix are the eigenvectors as coordinate vectors wrt to the standard basis. Hence we apply $P $ to get a vector wrt standard basis, so that $A $ can be applied to it (if $A $ is standard matrix representation of our linear transfirmation). – AnyAD Jun 09 '19 at 22:36
  • @AnyAD As Señor Wences used to say, “Easy for you; hard for me.” $A\mathbf v=\lambda\mathbf v$ is ingrained in stone. The rest is an almost trivial symbolic manipulation. For other things, remembering “input” and “output” bases is indeed easy. – amd Jun 09 '19 at 22:43
  • Don't disagree. (Thinking about transition matrices is sometimes useful, obviously depending on what one is trying to highlight) – AnyAD Jun 09 '19 at 22:47