Why Diagonalization Simplifies Powers
Repeatedly multiplying a matrix by itself can quickly make its entries large and its arithmetic cumbersome. For a real symmetric matrix, orthogonal diagonalization provides a more direct route: in suitable orthonormal coordinates, the matrix acts by multiplying each coordinate by an eigenvalue. Repeating that action raises each eigenvalue to the corresponding power. The same idea also tells us exactly when negative powers exist: no eigenvalue can be zero.
Let \(\mathbf{A}\in\mathbb{R}^{n\times n}\) be symmetric. The Spectral Theorem says that there is an orthogonal matrix \(\mathbf{Q}\) and a real diagonal matrix \(\mathbf{D}\) such that \(\mathbf{A}=\mathbf{Q}\mathbf{D}\mathbf{Q}^{\top}\). Orthogonal means \(\mathbf{Q}^{\top}\mathbf{Q}=\mathbf{Q}\mathbf{Q}^{\top}=\mathbf{I}_n\). Write \(\mathbf{D}=\operatorname{diag}(\lambda_1,\dots,\lambda_n)\); the diagonal entries are the eigenvalues, listed in the same order as the corresponding columns of \(\mathbf{Q}\).
For positive powers, the diagonal entries of \(\mathbf{D}\) are simply raised to the power. For the zero power, the diagonal matrix becomes the identity. For a negative power, each eigenvalue must be nonzero, and its reciprocal is raised to the positive power. The theorem below makes these statements precise and establishes the condition for negative powers.
Proof. First consider \(k\geq1\). If the formula holds for \(k\), then, using \(\mathbf{Q}^{\top}\mathbf{Q}=\mathbf{I}_n\), $$ \mathbf{A}^{k+1}=\mathbf{A}^k\mathbf{A} =\mathbf{Q}\mathbf{D}^k\mathbf{Q}^{\top}\mathbf{Q}\mathbf{D}\mathbf{Q}^{\top} =\mathbf{Q}\mathbf{D}^{k+1}\mathbf{Q}^{\top}. $$ The formula for \(k=1\) is the given diagonalization, so induction proves it for every positive integer. For \(k=0\), \(\mathbf{Q}\mathbf{I}_n\mathbf{Q}^{\top}=\mathbf{I}_n\), which proves the stated convention.
Now suppose every \(\lambda_i\ne0\), so that \(\mathbf{D}^{-1}=\operatorname{diag}(\lambda_1^{-1},\dots,\lambda_n^{-1})\) is defined. Set \(\mathbf{B}=\mathbf{Q}\mathbf{D}^{-1}\mathbf{Q}^{\top}\). Then $$ \mathbf{A}\mathbf{B}=\mathbf{Q}\mathbf{D}\mathbf{D}^{-1}\mathbf{Q}^{\top}=\mathbf{I}_n, \qquad \mathbf{B}\mathbf{A}=\mathbf{Q}\mathbf{D}^{-1}\mathbf{D}\mathbf{Q}^{\top}=\mathbf{I}_n. $$ Thus \(\mathbf{A}^{-1}=\mathbf{B}\). Applying the positive-power formula to \(\mathbf{A}^{-1}\) gives the claimed expression for \(\mathbf{A}^{-m}\). Conversely, if some \(\lambda_i=0\), let \(\mathbf{e}_i\) be the \(i\)th standard basis vector and set \(\mathbf{v}=\mathbf{Q}\mathbf{e}_i\). Orthogonality gives \(\mathbf{v}\ne\mathbf{0}\), while \(\mathbf{A}\mathbf{v}=\mathbf{Q}\mathbf{D}\mathbf{e}_i=\mathbf{0}\). If an inverse existed, multiplying this equality by \(\mathbf{A}^{-1}\) would give \(\mathbf{v}=\mathbf{0}\), a contradiction. Therefore \(\mathbf{A}\) is not invertible, and its negative powers are undefined. \(\square\)
Computing Positive Powers
To use the formula, raise the diagonal entries of \(\mathbf{D}\) to the desired power and then multiply by \(\mathbf{Q}\) and \(\mathbf{Q}^{\top}\). The diagonalization does not need to be recomputed for each exponent. The following example compares the result with direct multiplication.
Worked Example: A Third Power by Diagonalization and Multiplication
Let $$ \mathbf{Q}=\frac{1}{\sqrt{2}}\begin{bmatrix}1&1\\1&-1\end{bmatrix}, \qquad \mathbf{D}=\begin{bmatrix}1&0\\0&2\end{bmatrix}, \qquad \mathbf{A}=\mathbf{Q}\mathbf{D}\mathbf{Q}^{\top}. $$ The columns of \(\mathbf{Q}\) are orthonormal, so \(\mathbf{Q}^{\top}\mathbf{Q}=\mathbf{I}_2\). Multiplying out the diagonalization gives $$ \mathbf{A}=\frac12 \begin{bmatrix}1&1\\1&-1\end{bmatrix} \begin{bmatrix}1&1\\2&-2\end{bmatrix} =\begin{bmatrix}\frac32&-\frac12\\-\frac12&\frac32\end{bmatrix}. $$ The power formula yields $$ \mathbf{A}^3=\mathbf{Q} \begin{bmatrix}1&0\\0&8\end{bmatrix} \mathbf{Q}^{\top} =\begin{bmatrix}\frac92&-\frac72\\-\frac72&\frac92\end{bmatrix}. $$ To check by direct multiplication, first calculate $$ \mathbf{A}^2= \begin{bmatrix}\frac32&-\frac12\\-\frac12&\frac32\end{bmatrix} \begin{bmatrix}\frac32&-\frac12\\-\frac12&\frac32\end{bmatrix} =\begin{bmatrix}\frac52&-\frac32\\-\frac32&\frac52\end{bmatrix}. $$ Then $$ \mathbf{A}^2\mathbf{A} =\begin{bmatrix}\frac52&-\frac32\\-\frac32&\frac52\end{bmatrix} \begin{bmatrix}\frac32&-\frac12\\-\frac12&\frac32\end{bmatrix} =\begin{bmatrix}\frac92&-\frac72\\-\frac72&\frac92\end{bmatrix}. $$ This agrees with the diagonalization result, and direct multiplication verifies the computed third power.
The zero exponent is a useful boundary case: even if \(\mathbf{A}\) has a zero eigenvalue, \(\mathbf{A}^0\) is still the identity by definition. In the diagonal formula, the middle matrix for this case is \(\mathbf{I}_n\), rather than a calculation involving \(0^0\). For a positive exponent, a zero eigenvalue remains zero, so the corresponding diagonal entry of \(\mathbf{A}^k\) is zero.
Computing Negative Powers
A negative power requires every eigenvalue to be nonzero. Once that condition holds, the calculation follows the same pattern as for positive powers, but the diagonal entries are reciprocals before they are raised to the desired exponent. A direct multiplication check is especially useful: for example, a proposed \(\mathbf{A}^{-2}\) should satisfy \(\mathbf{A}^2\mathbf{A}^{-2}=\mathbf{I}_n\).
Worked Example: A Negative Second Power
Use the same orthogonal matrix as above, but let $$ \mathbf{D}=\begin{bmatrix}2&0\\0&-1\end{bmatrix}, \qquad \mathbf{A}=\mathbf{Q}\mathbf{D}\mathbf{Q}^{\top} =\begin{bmatrix}\frac12&\frac32\\\frac32&\frac12\end{bmatrix}. $$ Both eigenvalues are nonzero, so negative powers are defined. The formula gives $$ \mathbf{A}^{-2}=\mathbf{Q} \begin{bmatrix}\frac14&0\\0&1\end{bmatrix} \mathbf{Q}^{\top} =\begin{bmatrix}\frac58&-\frac38\\-\frac38&\frac58\end{bmatrix}. $$ To check the computation, first calculate the inverse from the same diagonalization: $$ \mathbf{A}^{-1}=\mathbf{Q} \begin{bmatrix}\frac12&0\\0&-1\end{bmatrix} \mathbf{Q}^{\top} =\begin{bmatrix}-\frac14&\frac34\\\frac34&-\frac14\end{bmatrix}. $$ Squaring this matrix gives $$ \mathbf{A}^{-1}\mathbf{A}^{-1} =\begin{bmatrix}\frac1{16}+\frac9{16}&-\frac3{16}-\frac3{16}\\ -\frac3{16}-\frac3{16}&\frac9{16}+\frac1{16}\end{bmatrix} =\begin{bmatrix}\frac58&-\frac38\\-\frac38&\frac58\end{bmatrix}, $$ which agrees with the claimed \(\mathbf{A}^{-2}\). For a further check, direct multiplication gives $$ \mathbf{A}^2= \begin{bmatrix}\frac52&\frac32\\\frac32&\frac52\end{bmatrix}, \qquad \mathbf{A}^2\mathbf{A}^{-2} =\begin{bmatrix}1&0\\0&1\end{bmatrix}. $$ For instance, the upper-left entry of the product is \(\frac52\cdot\frac58+\frac32\cdot(-\frac38)=\frac{25}{16}-\frac9{16}=1\); the upper-right entry is \(\frac52\cdot(-\frac38)+\frac32\cdot\frac58=0\). The other two entries are \(0\) and \(1\) by the same displayed products.
Negative eigenvalues do not prevent negative powers. In the example, the eigenvalue \(-1\) contributes \((-1)^{-2}=1\) to the second negative power. Signs depend on the exponent: for an odd negative exponent, that same eigenvalue contributes \((-1)^{-1}=-1\). The restriction is nonzero eigenvalues, not positive eigenvalues.
When Negative Powers Do Not Exist
Worked Example: A Zero Eigenvalue Blocks Negative Powers
Let $$ \mathbf{Q}=\frac{1}{\sqrt{2}}\begin{bmatrix}1&1\\1&-1\end{bmatrix}, \qquad \mathbf{D}=\begin{bmatrix}3&0\\0&0\end{bmatrix}. $$ Then $$ \mathbf{A}=\mathbf{Q}\mathbf{D}\mathbf{Q}^{\top} =\begin{bmatrix}\frac32&\frac32\\\frac32&\frac32\end{bmatrix}. $$ The positive second power is well-defined. The formula gives $$ \mathbf{A}^2=\mathbf{Q} \begin{bmatrix}9&0\\0&0\end{bmatrix} \mathbf{Q}^{\top} =\begin{bmatrix}\frac92&\frac92\\\frac92&\frac92\end{bmatrix}. $$ Direct multiplication confirms every entry: $$ \mathbf{A}\mathbf{A} =\begin{bmatrix}\frac32\cdot\frac32+\frac32\cdot\frac32& \frac32\cdot\frac32+\frac32\cdot\frac32\\ \frac32\cdot\frac32+\frac32\cdot\frac32& \frac32\cdot\frac32+\frac32\cdot\frac32\end{bmatrix} =\begin{bmatrix}\frac92&\frac92\\\frac92&\frac92\end{bmatrix}. $$ But \(\mathbf{A}\begin{bmatrix}1\\-1\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}\), even though \(\begin{bmatrix}1\\-1\end{bmatrix}\ne\mathbf{0}\). Thus \(\mathbf{A}\) has no inverse, and \(\mathbf{A}^{-1}\), and hence every negative integer power, is undefined.
Common Pitfalls and a Reliable Procedure
Common mistake: taking reciprocals of a zero eigenvalue. For the last example, one might try to write \(\mathbf{D}^{-1}=\begin{bmatrix}\frac13&0\\0&\frac10\end{bmatrix}\). The second entry is undefined, so this is not a matrix and cannot be used to form \(\mathbf{A}^{-1}\). The correction is to check all diagonal entries of \(\mathbf{D}\) first: if even one is zero, stop—negative powers do not exist. Positive powers remain valid.
Common mistake: confusing the zero matrix power with a zero matrix. The definition gives \(\mathbf{A}^0=\mathbf{I}_n\), not \(\mathbf{O}\). This holds whether or not \(\mathbf{A}\) is invertible. In the diagonal formula, use \(\mathbf{I}_n\) for the middle matrix at exponent zero; do not attempt to evaluate \(0^0\).
For an actual calculation, use the following sequence.
Write \(\mathbf{A}=\mathbf{Q}\mathbf{D}\mathbf{Q}^{\top}\) and confirm that \(\mathbf{Q}^{\top}\mathbf{Q}=\mathbf{I}_n\).
For a positive exponent, raise each diagonal entry to that power. For exponent zero, use \(\mathbf{I}_n\). For a negative exponent, first verify that every diagonal entry is nonzero.
Place the powered diagonal entries between \(\mathbf{Q}\) and \(\mathbf{Q}^{\top}\), then multiply if an explicit matrix is required.
For a positive power, multiply the required number of copies of \(\mathbf{A}\) on a small example. For a negative power, check the corresponding product with the positive power gives \(\mathbf{I}_n\).
This method matters because it separates the repeated action into independent scalar calculations. Instead of repeatedly multiplying a full matrix, one works with its eigenvalues and then returns to the original coordinates. The condition for negative powers is also transparent: a zero eigenvalue makes one direction collapse to \(\mathbf{0}\), so there can be no inverse that recovers that direction. Repeated eigenvalues cause no difficulty; each diagonal entry is powered according to its value, regardless of whether another entry is equal to it.
Check Your Understanding
Use the diagonalization formula, and where requested verify the result by direct multiplication.
- If \(\mathbf{A}=\mathbf{Q}\operatorname{diag}(2,3)\mathbf{Q}^{\top}\) for an orthogonal \(\mathbf{Q}\), what diagonal matrix appears in the formula for \(\mathbf{A}^4\)?
- For \(\mathbf{A}=\mathbf{Q}\operatorname{diag}(2,-1)\mathbf{Q}^{\top}\), compute the diagonal matrix used to find \(\mathbf{A}^{-3}\).
- Can \(\mathbf{A}^{-1}\) be defined if \(\mathbf{A}=\mathbf{Q}\operatorname{diag}(4,0,1)\mathbf{Q}^{\top}\)? Give the reason.
- For \(\mathbf{A}=\mathbf{Q}\operatorname{diag}(3,0)\mathbf{Q}^{\top}\), state the diagonal matrix in the formula for \(\mathbf{A}^0\).
- For the matrix in Worked Example: A Negative Second Power, verify that \(\mathbf{A}\mathbf{A}^{-1}=\mathbf{I}_2\) using the displayed matrices.
