Diagonalisation

≈ 25 min

A square matrix $A$ is diagonalisable if it can be written

A = P D P^{-1},

where $D$ is diagonal (its entries are the eigenvalues of $A$ ) and the columns of $P$ are the corresponding eigenvectors. This is possible exactly when $A$ has a full set of $n$ linearly independent eigenvectors.

Why it is useful. Powers become trivial:

A^k = P D^k P^{-1},

and $D^k$ just raises each diagonal entry to the $k$ -th power. Computing $A^{10}$ directly is punishing; through $D$ it is one line.

Worked example. $A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$ has eigenvalues $3, 1$ with eigenvectors $(1,1), (1,-1)$ , so $P = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$ and $D = \begin{pmatrix} 3 & 0 \\ 0 & 1 \end{pmatrix}.$

An $n \times n$ matrix is diagonalisable exactly when it has…

$n$ linearly independent eigenvectors

a determinant of zero

all positive entries

at least one eigenvalue equal to 1

$A$ has eigenvalues $3$ and $1$ . What is the largest eigenvalue of $A^2$ ?

In the factorisation $A = P D P^{-1}$ , the columns of $P$ are…

eigenvectors of $A$

the rows of $A$

the eigenvalues of $A$

always the standard basis vectors

$A$ has eigenvalues $2$ and $3$ . Find the trace of $A^3$ (the sum of the eigenvalues of $A^3$ ).