Proof: Powers of a Diagonalizable Matrix

Why Diagonalization Simplifies Powers

Repeatedly multiplying a matrix by itself can become cumbersome, even when the matrix has a simple structure hidden by its entries. A diagonalization expresses that structure as a change of coordinates: the diagonal matrix records how the matrix acts along eigenvector directions, while the invertible matrix \(P\) changes between those directions and the original coordinates. This tutorial proves that the change of coordinates can be kept outside any nonnegative integer power.

Throughout, matrices and vectors have real entries. Let \(\mathbf{A}\in\mathbb{R}^{n\times n}\). For an integer \(k\ge 1\), define \(\mathbf{A}^k\) to be the product of \(k\) copies of \(\mathbf{A}\), and define \(\mathbf{A}^0=\mathbf{I}_n\). A matrix is diagonalizable over \(\mathbb{R}\) if it has a representation \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\), where \(\mathbf{P}\in\mathbb{R}^{n\times n}\) is invertible and \(\mathbf{D}\in\mathbb{R}^{n\times n}\) is diagonal. In the usual eigenvector interpretation, the columns of \(\mathbf{P}\) are a basis of eigenvectors and the corresponding eigenvalues appear in the same order along the diagonal of \(\mathbf{D}\).

The theorem concerns nonnegative integer powers. It does not require \(\mathbf{A}\) to be invertible: \(\mathbf{P}\) must be invertible, but diagonal entries of \(\mathbf{D}\) may be zero. The exponent-zero case is included, and it is important because the definition gives \(\mathbf{A}^0=\mathbf{I}_n\), not a zero matrix.

Proof by Induction

Proof. Fix an integer \(k\ge 0\) and use induction. For the base case \(k=0\), the definitions of zeroth powers and the inverse give \(\mathbf{A}^0=\mathbf{I}_n=\mathbf{P}\mathbf{I}_n\mathbf{P}^{-1}=\mathbf{P}\mathbf{D}^0\mathbf{P}^{-1}\). Now suppose, for some \(k\ge 0\), that \(\mathbf{A}^k=\mathbf{P}\mathbf{D}^k\mathbf{P}^{-1}\). Then, using the definition of the next power and associativity of matrix multiplication, \(\mathbf{A}^{k+1}=\mathbf{A}^k\mathbf{A}=(\mathbf{P}\mathbf{D}^k\mathbf{P}^{-1})(\mathbf{P}\mathbf{D}\mathbf{P}^{-1})=\mathbf{P}\mathbf{D}^k(\mathbf{P}^{-1}\mathbf{P})\mathbf{D}\mathbf{P}^{-1}=\mathbf{P}\mathbf{D}^k\mathbf{D}\mathbf{P}^{-1}=\mathbf{P}\mathbf{D}^{k+1}\mathbf{P}^{-1}\). The cancellation is valid because \(\mathbf{P}^{-1}\mathbf{P}=\mathbf{I}_n\), and multiplication by the identity leaves a matrix unchanged. Thus the statement holds at \(k=0\), and its truth at any \(k\ge0\) implies its truth at \(k+1\). By induction, it holds for every integer \(k\ge0\). \(\square\)

The proof relies on the neighboring factors \(\mathbf{P}^{-1}\mathbf{P}\) becoming the identity at each multiplication step. It does not move the factors past one another: matrix multiplication is generally not commutative. The diagonal factors combine as \(\mathbf{D}^k\mathbf{D}=\mathbf{D}^{k+1}\) because they are consecutive copies of the same matrix.

What the Diagonal Power Means

For a diagonal matrix, taking a nonnegative integer power acts entry by entry. The convention for the zeroth power is that every diagonal entry becomes \(1\), including an entry that was \(0\). Thus, if \(\mathbf{D}=\operatorname{diag}(d_1,\dots,d_n)\), then \(\mathbf{D}^k=\operatorname{diag}(d_1^k,\dots,d_n^k)\) for \(k\ge1\), while \(\mathbf{D}^0=\mathbf{I}_n\). Here \(\operatorname{diag}(d_1,\dots,d_n)\) denotes the diagonal matrix with those diagonal entries in that order.

Proof. For \(k=0\), the definition gives \(\mathbf{D}^0=\mathbf{I}_n\), which is diagonal with every diagonal entry equal to \(1\). Now let \(k\ge1\). A product of two diagonal matrices \(\operatorname{diag}(x_1,\dots,x_n)\) and \(\operatorname{diag}(y_1,\dots,y_n)\) is \(\operatorname{diag}(x_1y_1,\dots,x_ny_n)\): in row \(i\), the only potentially nonzero entry in the first matrix is in column \(i\), and the only potentially nonzero entry in column \(j\) of the second is in row \(j\), so the product has zero entries when \(i\ne j\) and entry \(x_i y_i\) when \(i=j\). Starting from \(\mathbf{D}\), this multiplication rule shows by induction on \(k\) that each successive product remains diagonal and multiplies each diagonal entry by itself once more. Its entries are therefore \(d_i^k\), as claimed. \(\square\)

This proposition explains why diagonalization is useful for powers: the matrix multiplication in the diagonal coordinates reduces to separate scalar powers. It also makes explicit that the eigenvalue \(0\) causes no difficulty for \(k\ge1\), since \(0^k=0\). For \(k=0\), one uses the identity matrix by definition rather than applying a scalar-power rule at zero.

The Exponent-Zero Case and Eigenvectors

The same representation also describes what powers do to the eigenvectors that form the columns of \(\mathbf{P}\). Write \(\mathbf{P}=\begin{bmatrix}\mathbf{v}_1&\cdots&\mathbf{v}_n\end{bmatrix}\) and \(\mathbf{D}=\operatorname{diag}(d_1,\dots,d_n)\). From \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\), the \(i\)-th columns on the two sides give \(\mathbf{A}\mathbf{v}_i=d_i\mathbf{v}_i\). Thus \(d_i\) is the eigenvalue corresponding to the nonzero eigenvector \(\mathbf{v}_i\).

Proof. For \(k=0\), \(\mathbf{A}^0\mathbf{v}=\mathbf{I}_n\mathbf{v}=\mathbf{v}\), which is the stated zeroth-power case. Suppose the claim holds for some \(k\ge0\). Then \(\mathbf{A}^{k+1}\mathbf{v}=\mathbf{A}(\mathbf{A}^k\mathbf{v})=\mathbf{A}(\lambda^k\mathbf{v})=\lambda^k\mathbf{A}\mathbf{v}=\lambda^k(\lambda\mathbf{v})=\lambda^{k+1}\mathbf{v}\). The induction proves the assertion for every nonnegative integer \(k\). \(\square\)

Common Mistake: Moving Factors Past One Another

A tempting but incorrect step is to treat each factor in \(\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\) as though it can be powered independently. For the matrices in the first example, this would give the false rule \((\mathbf{P}\mathbf{D}\mathbf{P}^{-1})^2=\mathbf{P}^2\mathbf{D}^2(\mathbf{P}^{-1})^2\). In fact, \(\mathbf{P}^2=\begin{bmatrix}1&2\\0&1\end{bmatrix}\) and \((\mathbf{P}^{-1})^2=\begin{bmatrix}1&-2\\0&1\end{bmatrix}\), so that incorrect expression evaluates to

This disagrees with the correct square \(\begin{bmatrix}4&5\\0&9\end{bmatrix}\). The correction is to preserve the order of the factors and group only adjacent terms: \((\mathbf{P}\mathbf{D}\mathbf{P}^{-1})(\mathbf{P}\mathbf{D}\mathbf{P}^{-1})=\mathbf{P}\mathbf{D}(\mathbf{P}^{-1}\mathbf{P})\mathbf{D}\mathbf{P}^{-1}=\mathbf{P}\mathbf{D}^2\mathbf{P}^{-1}\). The inverse factors cancel because they are adjacent, not because matrix factors may be rearranged.

Another important distinction is between the assumptions on \(\mathbf{P}\) and \(\mathbf{A}\). The proof uses the inverse of \(\mathbf{P}\), so \(\mathbf{P}\) must be invertible. It does not use an inverse of \(\mathbf{A}\); therefore, the theorem remains valid when \(\mathbf{A}\) is singular, as the example with a zero diagonal entry demonstrates. The formula is a consequence of the given diagonalization, not a claim that every square matrix has one.