4

When reading eigenvector of a matrix, there is a formula:

$AP = PD$

where in $P$, each column is A's eigenvector and $D$ is diagonal matrix with diagonal element being A's eigen values.

Now coming the question:

Is matrix P always invertible? because I often see equation $A = PDP^{-1}$ here and there, but don't know how to proof.

user2262504
  • 954

3 Answers3

2

If $A$ is diagonalizable, then yes. In some (all? most?) texts the very definition of an $n\times n$ matrix $A$ being diagonalizable over a field $F$ (let's assume $\mathbb{R}$) is that there exists a basis of $\mathbb{R}^n$ made from the eigenvectors of $A$.

The columns of $P$ are exactly these eigenvectors, and them being a basis implies their linear independence. Hence $P$ is an invertible matrix.

David P
  • 12,320
1

The theorem you appear to be using is the spectral theorem. It works for real symmetric matrices, but not for real asymmetric matrices (nor indeed for matrices that aren't square). For that you need the singular value decomposition.

For all such issues I'd recommend Gilbert Strang's linear algebra book.

JPi
  • 4,562
0

There is a theorem that states that eigenvectors $v_1, v_2, \dotsc, v_k$ corresponding to distinct eigenvalues $\lambda_1, \lambda_2, \dotsc, \lambda_k$ are always linearly independent (reference).

If we're in $\mathbb{R}^n$, then the $k$ linearly independent eigenvectors will form a basis for $\mathbb{R}^n$ if $k$ = $n$ (reference). Therefore, we need the algebraic and geometric multiplicities for each eigenvalue to be equal (which is also the condition for $A$ being diagonalizable). The algebraic multiplicity always sum to $n$, and we need $n$ linearly independent eigenvectors (given by the geometric multiplicity). Also, if $n \neq k$, then $P$ is not square, so it won't be invertible.

The $n$ linearly independent eigenvectors will form a basis $\mathcal{B}$. The matrix $P$ is thus the change-of-basis matrix from the standard basis to $\mathcal{B}$. The change-of-basis matrix always has an inverse (proof).

Therefore $P$ is invertible $\Leftrightarrow$ $A$ is diagonalizable $\Leftrightarrow$ algebraic multiplicity = geometric multiplicity $\Leftrightarrow$ $A$ has $n$ distinct eigenvalues ($k = n$) $\Leftrightarrow$ $P$ is square.

aiwl
  • 181