From Diagonalization to Negative Powers
Diagonalization simplifies positive powers because powers of a diagonal matrix are found by taking powers of its diagonal entries. Negative powers follow the same pattern, but only when the matrix is invertible: each diagonal entry must have a reciprocal. This condition is essential, not a technical detail. If a diagonal entry is zero, the corresponding eigenvector is sent to the zero vector, and no inverse matrix can undo that action.
Throughout this tutorial, matrices and vectors have real entries. For a square matrix \(\mathbf{A}\), a negative integer power is defined only when \(\mathbf{A}\) is invertible. For an integer \(k\geq 1\), set \(\mathbf{A}^{-k}=(\mathbf{A}^{-1})^k\). Thus \(\mathbf{A}^{-2}=\mathbf{A}^{-1}\mathbf{A}^{-1}\), and \(\mathbf{A}^{-3}=\mathbf{A}^{-1}\mathbf{A}^{-1}\mathbf{A}^{-1}\). This definition does not permit \(\mathbf{A}^{-k}\) when \(\mathbf{A}^{-1}\) does not exist.
For example, if \(\mathbf{D}=\operatorname{diag}(\lambda_1,\dots,\lambda_n)\), then multiplying diagonal matrices gives \(\mathbf{D}^k=\operatorname{diag}(\lambda_1^k,\dots,\lambda_n^k)\) for positive integers \(k\). For a negative power, the analogous entrywise operation is possible only if none of the \(\lambda_i\) is zero. The next result justifies both that condition and the order of the matrix factors.
Proof. First suppose every \(\lambda_i\ne0\). Define \(\mathbf{D}^{-1}=\operatorname{diag}(\lambda_1^{-1},\dots,\lambda_n^{-1})\). The diagonal products \(\mathbf{D}\mathbf{D}^{-1}\) and \(\mathbf{D}^{-1}\mathbf{D}\) are both \(\mathbf{I}_n\). Therefore $$ \mathbf{A}\bigl(\mathbf{P}\mathbf{D}^{-1}\mathbf{P}^{-1}\bigr) =\mathbf{P}\mathbf{D}\mathbf{D}^{-1}\mathbf{P}^{-1} =\mathbf{I}_n, $$ and $$ \bigl(\mathbf{P}\mathbf{D}^{-1}\mathbf{P}^{-1}\bigr)\mathbf{A} =\mathbf{P}\mathbf{D}^{-1}\mathbf{D}\mathbf{P}^{-1} =\mathbf{I}_n. $$ Thus \(\mathbf{A}\) is invertible and \(\mathbf{A}^{-1}=\mathbf{P}\mathbf{D}^{-1}\mathbf{P}^{-1}\). For any \(k\geq1\), multiplying \(k\) copies of this expression makes each adjacent \(\mathbf{P}^{-1}\mathbf{P}\) equal \(\mathbf{I}_n\). Hence \((\mathbf{A}^{-1})^k=\mathbf{P}(\mathbf{D}^{-1})^k\mathbf{P}^{-1}\). The middle matrix is diagonal with entries \(\lambda_i^{-k}\), proving the formula. Conversely, if some \(\lambda_j=0\), let \(\mathbf{v}\) be column \(j\) of \(\mathbf{P}\). Since \(\mathbf{P}\) is invertible, \(\mathbf{v}\ne\mathbf{0}\), and \(\mathbf{A}\mathbf{v}=\lambda_j\mathbf{v}=\mathbf{0}\). If \(\mathbf{A}\) had an inverse, applying it to this equality would give \(\mathbf{v}=\mathbf{0}\), a contradiction. Therefore \(\mathbf{A}\) is not invertible. \(\square\)
Calculating and Checking an Inverse Power
The theorem is a procedure: find or use a diagonalization, replace every diagonal eigenvalue by its negative power, and keep \(\mathbf{P}\) and \(\mathbf{P}^{-1}\) in their original positions. This last point matters because matrix multiplication is not generally commutative. A useful check is to multiply the proposed inverse by the original matrix, or, for a higher negative power, to multiply by the corresponding positive power.
Worked Example: Find the Inverse and the Inverse Square
Let $$ \mathbf{P}=\begin{bmatrix}1&1\\0&1\end{bmatrix}, \qquad \mathbf{D}=\begin{bmatrix}2&0\\0&4\end{bmatrix}, \qquad \mathbf{P}^{-1}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}. $$ The claimed inverse of \(\mathbf{P}\) checks because \(\mathbf{P}\mathbf{P}^{-1}=\mathbf{P}^{-1}\mathbf{P}=\mathbf{I}_2\). Define \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\). First, $$ \mathbf{P}\mathbf{D} =\begin{bmatrix}2&4\\0&4\end{bmatrix}, \qquad \mathbf{A} =\begin{bmatrix}2&4\\0&4\end{bmatrix} \begin{bmatrix}1&-1\\0&1\end{bmatrix} =\begin{bmatrix}2&2\\0&4\end{bmatrix}. $$ Both diagonal entries of \(\mathbf{D}\) are nonzero. The formula gives $$ \mathbf{A}^{-1} =\mathbf{P}\begin{bmatrix}\frac12&0\\0&\frac14\end{bmatrix}\mathbf{P}^{-1} =\begin{bmatrix}\frac12&\frac14\\0&\frac14\end{bmatrix} \begin{bmatrix}1&-1\\0&1\end{bmatrix} =\begin{bmatrix}\frac12&-\frac14\\0&\frac14\end{bmatrix}. $$ Multiplying back verifies the answer: $$ \mathbf{A}\mathbf{A}^{-1} =\begin{bmatrix}2&2\\0&4\end{bmatrix} \begin{bmatrix}\frac12&-\frac14\\0&\frac14\end{bmatrix} =\begin{bmatrix}1&0\\0&1\end{bmatrix}. $$ For the inverse square, take the reciprocal squares \(2^{-2}=\frac14\) and \(4^{-2}=\frac1{16}\): $$ \mathbf{A}^{-2} =\mathbf{P}\begin{bmatrix}\frac14&0\\0&\frac1{16}\end{bmatrix}\mathbf{P}^{-1} =\begin{bmatrix}\frac14&\frac1{16}\\0&\frac1{16}\end{bmatrix} \begin{bmatrix}1&-1\\0&1\end{bmatrix} =\begin{bmatrix}\frac14&-\frac3{16}\\0&\frac1{16}\end{bmatrix}. $$ Here \(\mathbf{A}^2=\begin{bmatrix}4&12\\0&16\end{bmatrix}\), so $$ \mathbf{A}^2\mathbf{A}^{-2} =\begin{bmatrix}4&12\\0&16\end{bmatrix} \begin{bmatrix}\frac14&-\frac3{16}\\0&\frac1{16}\end{bmatrix} =\begin{bmatrix}1&0\\0&1\end{bmatrix}. $$ This verifies the inverse-square calculation as well.
The eigenvalues may be negative, too. Their reciprocal powers retain the usual sign: for instance, \((-2)^{-3}=-\frac18\), whereas \((-2)^{-2}=\frac14\). The eigenvector matrix combines these diagonal actions back into the original coordinates; negative powers do not mean that the entries of \(\mathbf{A}\) themselves are raised to negative powers.
Worked Example: Negative Eigenvalues and an Odd Inverse Power
Consider $$ \mathbf{P}=\begin{bmatrix}1&1\\1&-1\end{bmatrix}, \qquad \mathbf{P}^{-1}=\frac12\begin{bmatrix}1&1\\1&-1\end{bmatrix}, \qquad \mathbf{D}=\begin{bmatrix}2&0\\0&-1\end{bmatrix}. $$ Direct multiplication gives \(\mathbf{P}\mathbf{P}^{-1}=\mathbf{I}_2\). The corresponding matrix is $$ \mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1} =\begin{bmatrix}2&-1\\2&1\end{bmatrix} \frac12\begin{bmatrix}1&1\\1&-1\end{bmatrix} =\begin{bmatrix}\frac12&\frac32\\[2pt]\frac32&\frac12\end{bmatrix}. $$ For the inverse cube, the diagonal entries become \(2^{-3}=\frac18\) and \((-1)^{-3}=-1\): $$ \mathbf{A}^{-3} =\mathbf{P}\begin{bmatrix}\frac18&0\\0&-1\end{bmatrix}\mathbf{P}^{-1} =\begin{bmatrix}\frac18&-1\\[2pt]\frac18&1\end{bmatrix} \frac12\begin{bmatrix}1&1\\1&-1\end{bmatrix} =\begin{bmatrix}-\frac7{16}&\frac9{16}\\[2pt]\frac9{16}&-\frac7{16}\end{bmatrix}. $$ To check it, the same diagonalization gives $$ \mathbf{A}^3 =\mathbf{P}\begin{bmatrix}8&0\\0&-1\end{bmatrix}\mathbf{P}^{-1} =\begin{bmatrix}\frac72&\frac92\\[2pt]\frac92&\frac72\end{bmatrix}. $$ Therefore $$ \mathbf{A}^3\mathbf{A}^{-3} =\frac1{32}\begin{bmatrix}-49+81&63-63\\63-63&81-49\end{bmatrix} =\mathbf{I}_2. $$ In particular, the negative eigenvalue contributes a negative diagonal entry to this odd inverse power.
What Happens When a Diagonal Entry Is Zero?
Diagonalizability alone does not guarantee that negative powers exist. A zero diagonal entry means there is a nonzero eigenvector \(\mathbf{v}\) with \(\mathbf{A}\mathbf{v}=\mathbf{0}\). Such a matrix cannot have an inverse, so writing a reciprocal of that zero entry is not a valid computation. The next example makes the distinction explicit.
Worked Example: A Diagonalizable Matrix Without Negative Powers
Let $$ \mathbf{A}=\begin{bmatrix}3&0\\0&0\end{bmatrix}. $$ This matrix is already diagonal, so it is diagonalizable with \(\mathbf{P}=\mathbf{I}_2\). But for \(\mathbf{v}=\begin{bmatrix}0\\1\end{bmatrix}\ne\mathbf{0}\), direct multiplication gives $$ \mathbf{A}\mathbf{v} =\begin{bmatrix}3&0\\0&0\end{bmatrix} \begin{bmatrix}0\\1\end{bmatrix} =\begin{bmatrix}0\\0\end{bmatrix}. $$ If an inverse existed, multiplying \(\mathbf{A}\mathbf{v}=\mathbf{0}\) by \(\mathbf{A}^{-1}\) would imply \(\mathbf{v}=\mathbf{0}\), contrary to its displayed value. Thus \(\mathbf{A}^{-1}\), and hence \(\mathbf{A}^{-k}\) for \(k\geq1\), is not defined. The tempting diagonal expression \(\operatorname{diag}(3^{-1},0^{-1})\) is invalid because \(0^{-1}\) does not exist.
The Eigenvectors Keep Their Roles
The formula also tells us what a negative power does to each eigenvector. This is useful as a conceptual check: an eigenvector does not change direction under the map; its scaling factor becomes the corresponding reciprocal power. Here is the precise statement, including why its hypothesis requires a nonzero eigenvalue.
Proof. If \(\lambda=0\), then \(\mathbf{A}\mathbf{v}=\mathbf{0}\); multiplying by the inverse would yield \(\mathbf{v}=\mathbf{0}\), a contradiction. Thus \(\lambda\ne0\). From \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\), multiply by \(\mathbf{A}^{-1}\) to obtain \(\mathbf{v}=\lambda\mathbf{A}^{-1}\mathbf{v}\). Dividing by the nonzero scalar \(\lambda\) gives \(\mathbf{A}^{-1}\mathbf{v}=\lambda^{-1}\mathbf{v}\). Repeating this relation \(k\) times gives \((\mathbf{A}^{-1})^k\mathbf{v}=\lambda^{-k}\mathbf{v}\), which is the asserted formula. \(\square\)
Common Mistake: Reversing the Matrix Factors
A frequent error is to write \(\mathbf{A}^{-1}=\mathbf{P}^{-1}\mathbf{D}^{-1}\mathbf{P}\). The factors must instead be \(\mathbf{P}\mathbf{D}^{-1}\mathbf{P}^{-1}\). The order can be checked by multiplying against \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\): the middle factors \(\mathbf{P}^{-1}\mathbf{P}\) then cancel to \(\mathbf{I}_n\).
In the first example, the incorrect order would give $$ \mathbf{P}^{-1}\mathbf{D}^{-1}\mathbf{P} =\begin{bmatrix}\frac12&-\frac14\\0&\frac14\end{bmatrix} \begin{bmatrix}1&1\\0&1\end{bmatrix} =\begin{bmatrix}\frac12&\frac14\\0&\frac14\end{bmatrix}. $$ But multiplying this proposed answer by \(\mathbf{A}\) gives $$ \begin{bmatrix}2&2\\0&4\end{bmatrix} \begin{bmatrix}\frac12&\frac14\\0&\frac14\end{bmatrix} =\begin{bmatrix}1&1\\0&1\end{bmatrix} \ne\mathbf{I}_2. $$ The correction is to keep the original outer factors in place and replace only \(\mathbf{D}\) by \(\mathbf{D}^{-1}\). This cancellation-based check works for every invertible diagonalization, not just for this example.
When using the formula, first check that all diagonal entries are nonzero. Then replace each \(\lambda_i\) by \(\lambda_i^{-k}\), retain the order \(\mathbf{P}\), diagonal matrix, \(\mathbf{P}^{-1}\), and verify by multiplication when practical. A repeated diagonal entry causes no special rule: each occurrence is replaced by the same reciprocal power. The calculation depends on the actual diagonal entries and on the chosen matching eigenvector columns, not on rearranging the factors.
Check Your Understanding
For each question, identify the invertibility condition before calculating a negative power.
- If \(\mathbf{A}=\mathbf{P}\operatorname{diag}(5,-2)\mathbf{P}^{-1}\), write a formula for \(\mathbf{A}^{-2}\).
- Can a diagonalizable matrix with diagonal form \(\operatorname{diag}(4,0)\) have an inverse? Explain using a nonzero vector.
- For \(\mathbf{A}=\mathbf{P}\operatorname{diag}(-3,1)\mathbf{P}^{-1}\), what diagonal matrix appears in the formula for \(\mathbf{A}^{-3}\)?
- Explain why \(\mathbf{P}\mathbf{D}^{-1}\mathbf{P}^{-1}\) is the correct order by multiplying it by \(\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\).
- If \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\) with \(\mathbf{v}\ne\mathbf{0}\) and \(\mathbf{A}\) invertible, state \(\mathbf{A}^{-2}\mathbf{v}\) and why \(\lambda\) cannot be zero.
