Given diagonal matrix, is it possible to find the invertible matrix P?

Question

A is not explicitly given, but A satisfies the following.

$A\begin{bmatrix}1\\1\\0\end{bmatrix}=\begin{bmatrix}1\\1\\0\end{bmatrix},A\begin{bmatrix}1\\0\\1\end{bmatrix}=-\begin{bmatrix}1\\0\\1\end{bmatrix}, A\begin{bmatrix}0\\1\\1\end{bmatrix}=2\begin{bmatrix}0\\1\\1\end{bmatrix}$

Is it possible to find an invertible matrix $P$ such that $P^{-1}AP=D=\begin{bmatrix}1&0&0\\0&-1&0\\0&0&2\end{bmatrix}$

Are those matrices in the equations have something to do with $P$?

Are they colums of $P$? If yes, why?

Answer 1

I'll try to give more insight and intuition than other answers provided thus far. The key here is to have a very strong grasp of the following fact that is easy to verify by direct computation (of course it holds for general matrices but it's helpful to make it concrete here):

If $[v_1\mid v_2\mid v_3]$ are the columns of the $3\times 3$ matrix $A$, and $b$ is the column vector $(b_1, b_2, b_3)^T$ then the vector $Ab$ is equal to the linear combination $b_1 v_1 + b_2 v_2 + b_3 v_3$.

This has a wide range of consequences, one of them being that we can reveal the columns of $A$ by simply multiplying $A$ against each of the column vectors $(1,0,0)^T$, $(0,1,0)^T$ and $(0,0,1)^T$.

Suppose we form the matrix $P$ from the three column vectors in the question. It's easy to see that $P(1,0,0)^T = (1,1,0)^T$. Furthermore, (multiplying both sides on the left by $P^{-1}$) we have $(1,0,0)^T = P^{-1} (1,1,0)^T$.

[If you think about it, $P$ acts as a change of basis matrix converting from "basis-space" to the standard basis, and $P^{-1}$ is the reverse transformation, taking a vector and returning its components with respect to the basis.]

Now let's compute $P^{-1} A P$. By the prior observation we can compute this matrix one column at a time, by looking at say $x=(1,0,0)^T$ and computing $(P^{-1}AP)x$. By associativity this is the same as $P^{-1}(A(Px))$.

1. We know what $Px$ is: it's the first column of $P$, namely $(1,1,0)^T$.
2. Now we know what $A(Px) = A(1,1,0)^T$ is: it's exactly $(1,1,0)^T$, because that's one of the few clues we have about $A$.
3. Now we know what $P^{-1}((A(Px)) = P^{-1} (1,1,0)^T$ is: we just saw not too long ago that it's $(1,0,0)^T$.

So since we computed that $(P^{-1}AP)(1,0,0)^T = (1,0,0)^T$, we in fact know the first column of the matrix $P^{-1}AP$, and it indeed matches the first column of $D$.

Now, can you compute the other two columns of $P^{-1}AP$ using these ideas?

Answer 2

Note that, what appears form your given is that your matrix $A$ has eigen values $1$, $-1$, and $2$, and thier associatted eigen vectors are the given above vectors. ( Hope you are familiar with the notion of eigen values and eigen vectors)

According to Wolframe about diagonalisation of square matrices, the columns of matrix $P$ are the eigen vectors, ordered according to the ordering of thier associated eigen values in the diagonal matrix $D$. Infact this comes from the Eigen Decomposition of matrix $A$.

Answer 3

After reading the first statement, there is one thing that immediately tells you that $A$ is diagonalizable. Based on the hypothesis, $A$ is a $3\times 3$ matrix, and it scales $3$ different non-zero vectors by scalars $1$, $-1$, and $2$. By definition, these vectors are eigenvectors of $A$ corresponding to eigenvalues $1$, $-1$, and $2$ respectively. Since $A$ is a $3\times 3$ matrix and it has $3$ distinct eigenvalues, then $A$ is diagonalizable. (In more general terms, if $A$ is $n\times n$ and has $n$ distinct eigenvalues, then $A$ is diagonalizable.)

The eigenvectors of $A$ are used as column vectors to construct $P$. The diagonal form of $A$, namely, $D$, will contain the eigenvalues on the diagonal, and $D$ is unique up to reordering the entries on the diagonal. The entry $d_{ii}$ in $D$ will be the eigenvalue which corresponds to the eigenvector in the $i$th column of $P$.

The reason this works: If $A$ is $n\times n$ and diagonalizable, then $P^{-1}AP=D$ for some $P$, and diagonal matrix $D$. So then $AP=PD$. Then denote the $i$th column of $P$ as $\mathbf{p}_{i}$ and the $i$th diagonal entry of $D$ as $\lambda_{i}$. Observe: $$AP=A\begin{bmatrix}\mathbf{p}_{1} & \cdots & \mathbf{p}_{n}\end{bmatrix}=\begin{bmatrix}A\mathbf{p}_{1} & \cdots & A\mathbf{p}_{n}\end{bmatrix},$$ $$PD=\begin{bmatrix} | & & | \\ \mathbf{p}_{1} & \cdots & \mathbf{p}_{n} \\ | & & |\end{bmatrix}\begin{bmatrix}\lambda_{1} & & \\ & \ddots & \\ & & \lambda_{n}\end{bmatrix}=\begin{bmatrix} | & & | \\ \lambda_{1}\mathbf{p}_{1} & \cdots & \lambda_{n}\mathbf{p}_{n} \\ | & & |\end{bmatrix}.$$ So if $A$ is diagonalizable, we must have $A\mathbf{p}_{i}=\lambda_{i}\mathbf{p}_{i}$, for some $\lambda_{i}\in\mathbb{R}$ and each non-trivial vector $\mathbf{p}_{i}\in\mathbb{R}^{n}$ (since $\mathbf{p}_{i}$ is a column vector of an invertible matrix $P$, each $\mathbf{p}_{i}$ must be non trivial and linearly independent). Any $\mathbf{p}_{i}$ is, by definition, an eigenvector corresponding to eigenvalue $\lambda_{i}$. So we must use the eigenvectors to construct $P$.

If we have $n$ distinct eigenvalues, then $A$ will always be diagonalizable, because each eigenvalue will have at least one non-trivial eigenvector (since the characteristic polynomial of $A$ has each eigenvalue $\lambda_{i}$ as a root, and thus, $A-\lambda_{i}I$ must have a non-trivial solution). The reason that $P$ will be invertible has a slightly different proof, but for now, also note that a matrix $A$ will be invertible if and only if it has $n$ linearly-independent eigenvectors.