8. Diagonalizing a square matrix

For a square matrix $A$, to diagonalize $A$ means to find a square matrix $P$ and diagonal matrix $D$ with $P ^ {-1} A P = D$.

In this case we say that $P$ diagonalizes $A$. Notice that $P$ has to be nonsingular or it won't have an inverse.

Some matrices are ``diagonalizable'' and others are not.

$Example.$ For $A = \matp{cc}{4&1\\ 1&4}$, it turns out that $P = \matp{rr}{1&-1\\ 1&1}$ is a good matrix to diagonalize $A$. In fact, $P ^ {-1} A P = D$ for $D = \matp{cc}{5&0\\ 0&3}$.

Notice that $\matp{c}{1\\ 1}$ and $\matp{r}{-1\\ 1}$ are eigenvectors of $A$. In fact, $P$ is always made from eigenvectors:

Proposition. For $n \times n$ matrices $P$ and $A$, $P$ diagonalizes $A$ $\Leftrightarrow$ the columns of $P$ are linearly independent eigenvectors of $A$ and the diagonal entries of $D$ are the corresponding eigenvalues.

(The linear independence of columns is needed in order for $P$ to be nonsingular. It wouldn't work, for instance, to have all columns be the same. For the reasoning behind the Proposition, see below.)

The Proposition tells how to diagonalize $A$: Just find the eigenvalues and corresponding eigenvectors. If there are enough linearly independent eigenvectors, use them for the columns of $P$. The Example could be found that way.

Handy facts:

(1) Eigenvectors that belong to distinct eigenvalues are linearly independent. Therefore, if $A$ is $n \times n$ and has $n$ distinct eigenvalues, then $A$ is diagonalizable. (``Distinct'' means ``all different from one another''.)

(2) A real symmetric matrix can always be diagonalized, even if the eigenvalues are not all distinct (for example, if the characteristic polynomial factors as $(\lambda-1)(\lambda-1)(\lambda-4) (\lambda-6)$), we would say the eigenvalues are $1,1,4,6$, not all distinct.

These two facts will be proved in class. Let's go back and prove the Proposition.

Proof of the Proposition. For `` $\Rightarrow$'': To says that $P$ diagonalizes $A$ means that $P ^ {-1} A P = D$. If we take this equation and multiply both sides by $P$ on the left, we get $AP = PD$. Now let's try to interpret this second equation using eigenvalues and eigenvectors. We need two facts:

• First, whenever we multiply two matrices, such as $AP$, the columns of the matrix on the right work independently; if v$_ 1$ is the first column of $P$ then $A$v$_ 1$ is the first column of $AP$, and so on.

  For example, $\matp{rr}{1&2\\ 3&4}\matp{rr}{1&\cdot\\ 5&\cdot}= \matp{rr}{11&\cdot\\ 24& \cdot}$ and it doesn't matter what the other entries are.

• Second, whenever we multiply a matrix on the right by a diagonal matrix, as in $PD$, the effect is to multiply the columns of the left-hand matrix by scalars (the diagonal entries of $D$), so that if v$_ 1$ is the first column of $P$ then $d _ {11}$   v$_ 1$ is the first column of $PD$, and so on.

  For example, $\matp{rr}{1&2\\ 3&4}\matp{rr}{10&0\\ 0&100}= \matp{rr}{10&200\\ 30&400}$.

Putting these two facts together, we see that $A$v$_ 1 = d _ {11}$v$_ 1$, so v$_ 1$ is an eigenvector of $A$ and $d _ {11}$ is the corresponding eigenvalue. (An eigenvector can't be 0, but obviously v$_ 1 ne$   0 since it's a column of the invertible matrix $P$.) Now look at the second column v$_ 2$ of $P$; by the same facts we now get $A$v$_ 2 = d _ {22}$v$_ 2$, so v$_ 2$ is an eigenvector of $P$ with corresponding eigenvalue $d _ {22}$. The other columns of $P$ work similarly. The Proposition also mentions that the columns of $P$ are linearly independent, which is the same as saying that the column rank equals the number of columns, but this is automatic since $P$ is invertible.

We still need to prove `` $\Leftarrow$'' in the Proposition. We start by assuming we are given $A$ and $P$ such that the columns of $P$ are eigenvectors of $A$ and are linearly independent. The same two facts as before show that $AP = PD$. Since the columns of $P$ are linearly independent, $P$ has rank $n$ and is invertible. Multiply both sides of $AP = PD$ on the left by $P ^{-1}$; we get $P ^ {-1} A P = D$, as required.

Problem U-12. Diagonalize $A = \matp{rr}{7&-4\\ 2&1}$. (Use Problem U-.)

Problem U-13. Explain how to interpret the equation $P ^ {-1} A P = D$ in terms of a change of basis. What is the transformation? The old basis? The matrix relative to the old basis? The new basis? The transition matrix? The matrix relative to the new basis?