What Must Be Checked
A proposed diagonalization is a claim that a matrix \(\mathbf{A}\) can be written as \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\), where \(\mathbf{D}\) is diagonal and \(\mathbf{P}\) is invertible. Checking only that some entries of \(\mathbf{D}\) look like eigenvalues is not enough: the columns of \(\mathbf{P}\) must be eigenvectors matched, position by position, to the entries of \(\mathbf{D}\). The inverse in the formula also requires that \(\mathbf{P}\) actually be invertible.
Throughout, matrices and vectors have real entries. Suppose \(\mathbf{P}\in\mathbb{R}^{n\times n}\) has columns \(\mathbf{p}_1,\dots,\mathbf{p}_n\), and write the diagonal matrix as \(\mathbf{D}=\operatorname{diag}(\lambda_1,\dots,\lambda_n)\). Thus the \(j\)-th diagonal entry of \(\mathbf{D}\) is \(\lambda_j\). A vector \(\mathbf{p}\ne\mathbf{0}\) is an eigenvector of \(\mathbf{A}\) for the eigenvalue \(\lambda\) when \(\mathbf{A}\mathbf{p}=\lambda\mathbf{p}\).
A Shorter Equation for Verification
Computing the full product \(\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\) is one way to test a claim, but there is a shorter equivalent check. Multiplying the claimed factorization on the right by \(\mathbf{P}\) removes the inverse. This gives a matrix equation whose columns directly test the eigenvector matches.
(1) \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\).
(2) \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\).
(3) \(\mathbf{A}\mathbf{p}_j=\lambda_j\mathbf{p}_j\) for every \(j=1,\dots,n\).
Proof. Assume (1). Right-multiplying both sides by \(\mathbf{P}\) gives \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\mathbf{P}=\mathbf{P}\mathbf{D}\), because \(\mathbf{P}^{-1}\mathbf{P}=\mathbf{I}_n\). Thus (1) implies (2). Conversely, assume (2). Right-multiplying by \(\mathbf{P}^{-1}\) gives \(\mathbf{A}\mathbf{P}\mathbf{P}^{-1}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\), so \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\). Hence (1) and (2) are equivalent. For the equivalence of (2) and (3), the \(j\)-th column of \(\mathbf{A}\mathbf{P}\) is \(\mathbf{A}\mathbf{p}_j\). Since \(\mathbf{D}\) is diagonal, the \(j\)-th column of \(\mathbf{P}\mathbf{D}\) is \(\lambda_j\mathbf{p}_j\). Two matrices are equal exactly when their corresponding columns are equal, so \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\) exactly when \(\mathbf{A}\mathbf{p}_j=\lambda_j\mathbf{p}_j\) for every \(j\). \(\square\)
The invertibility hypothesis matters. It ensures that the expression \(\mathbf{P}^{-1}\) exists, and it also ensures that none of the columns of \(\mathbf{P}\) is zero. Indeed, if \(\mathbf{p}_j=\mathbf{0}\), then \(\mathbf{P}\mathbf{e}_j=\mathbf{0}\); multiplying by \(\mathbf{P}^{-1}\) would give \(\mathbf{e}_j=\mathbf{0}\), which is false. Therefore, when the column equations in (3) hold and \(\mathbf{P}\) is invertible, each column is a genuine eigenvector, not the zero vector.
Proof. If both \(\mathbf{A}\) and \(\mathbf{B}\) satisfy the stated factorization, then \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\) and \(\mathbf{B}\mathbf{P}=\mathbf{P}\mathbf{D}\) by the theorem. Thus \(\mathbf{A}\mathbf{P}=\mathbf{B}\mathbf{P}\). Right-multiplying by \(\mathbf{P}^{-1}\) gives \(\mathbf{A}=\mathbf{B}\). \(\square\)
Worked Checks
Worked Example: Verify a 2×2 Diagonalization
Check the proposed factorization for \(\mathbf{A}=\begin{bmatrix}2&1\\1&2\end{bmatrix}\), \(\mathbf{P}=\begin{bmatrix}1&1\\1&-1\end{bmatrix}\), and \(\mathbf{D}=\begin{bmatrix}3&0\\0&1\end{bmatrix}\).
First verify the proposed inverse. Let \(\mathbf{Q}=\frac12\begin{bmatrix}1&1\\1&-1\end{bmatrix}\). Multiplying in both orders gives
Therefore \(\mathbf{P}^{-1}=\mathbf{Q}\), and \(\mathbf{P}\) is invertible. Its first column is \(\mathbf{p}_1=(1,1)^{\top}\) and its second is \(\mathbf{p}_2=(1,-1)^{\top}\). Check each match:
The columns match the diagonal entries \(3,1\) in order. Equivalently, multiply the two sides of the shortcut equation:
They agree. To check the factorization by multiplying it out, first compute \(\mathbf{P}\mathbf{D}\), then multiply by the inverse:
Check: both the column equations and the full product give the stated matrix \(\mathbf{A}\), so the proposed diagonalization is valid.
Worked Example: Verify a 3×3 Diagonalization with a Repeated Entry
Check the claim for \(\mathbf{A}=\begin{bmatrix}2&0&0\\0&2&3\\0&0&5\end{bmatrix}\), \(\mathbf{P}=\begin{bmatrix}1&1&0\\0&1&1\\0&0&1\end{bmatrix}\), and \(\mathbf{D}=\begin{bmatrix}2&0&0\\0&2&0\\0&0&5\end{bmatrix}\). The repeated entry in \(\mathbf{D}\) does not change the checking procedure: each column still has its own position-specific match.
Use \(\mathbf{Q}=\begin{bmatrix}1&-1&1\\0&1&-1\\0&0&1\end{bmatrix}\) as the proposed inverse. Direct multiplication gives
Thus \(\mathbf{P}^{-1}=\mathbf{Q}\). The columns are \(\mathbf{p}_1=(1,0,0)^{\top}\), \(\mathbf{p}_2=(1,1,0)^{\top}\), and \(\mathbf{p}_3=(0,1,1)^{\top}\). Their corresponding diagonal entries are \(2,2,5\). Compute:
Now check the matrix equation. Multiplying each matrix gives
For the full product,
Check: the inverse products are identity matrices, each column has the required eigenvalue, and the full product returns \(\mathbf{A}\). All parts of the proposed factorization are verified.
Worked Example: An Ordering Error
Consider \(\mathbf{A}=\begin{bmatrix}2&0\\0&3\end{bmatrix}\), \(\mathbf{P}=\begin{bmatrix}1&0\\0&1\end{bmatrix}\), and the proposed \(\mathbf{D}=\begin{bmatrix}3&0\\0&2\end{bmatrix}\). The matrix \(\mathbf{P}\) is invertible, with \(\mathbf{P}^{-1}=\mathbf{I}_2\), but the order of the diagonal entries does not match the columns of \(\mathbf{P}\).
The first column is \(\mathbf{p}_1=(1,0)^{\top}\), and \(\mathbf{A}\mathbf{p}_1=(2,0)^{\top}\ne3\mathbf{p}_1\). The second column is \(\mathbf{p}_2=(0,1)^{\top}\), and \(\mathbf{A}\mathbf{p}_2=(0,3)^{\top}\ne2\mathbf{p}_2\). The shortcut check also fails:
Since \(\mathbf{P}^{-1}=\mathbf{I}_2\), the full product is \(\mathbf{P}\mathbf{D}\mathbf{P}^{-1}=\mathbf{D}\ne\mathbf{A}\). The proposed factorization is false. If we instead pair the first column with \(2\) and the second with \(3\), then \(\mathbf{D}=\operatorname{diag}(2,3)\) and the factorization holds.
When the Proposed Matrix Is Not Invertible
A list of eigenvectors that satisfies some of the expected equations is not, by itself, enough: the matrix of columns must be invertible. If \(\mathbf{P}\) is singular, then \(\mathbf{P}^{-1}\) does not exist, and the proposed expression \(\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\) is not defined. This is a separate check from matching columns to diagonal entries.
Worked Example: A Singular Eigenvector Matrix
Let \(\mathbf{A}=\begin{bmatrix}2&0\\0&3\end{bmatrix}\), \(\mathbf{P}=\begin{bmatrix}1&1\\0&0\end{bmatrix}\), and \(\mathbf{D}=\begin{bmatrix}2&0\\0&3\end{bmatrix}\). The columns of \(\mathbf{P}\) are both \((1,0)^{\top}\), so they are dependent. In particular, \(\mathbf{P}\begin{bmatrix}1\\-1\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}\), although \(\begin{bmatrix}1\\-1\end{bmatrix}\ne\mathbf{0}\). If an inverse \(\mathbf{P}^{-1}\) existed, multiplying this equation on the left by \(\mathbf{P}^{-1}\) would imply \(\begin{bmatrix}1\\-1\end{bmatrix}=\mathbf{0}\), a contradiction. Thus \(\mathbf{P}\) is not invertible.
Also, the second column cannot match the second diagonal entry: \(\mathbf{A}\begin{bmatrix}1\\0\end{bmatrix}=\begin{bmatrix}2\\0\end{bmatrix}\ne\begin{bmatrix}3\\0\end{bmatrix}\). The proposal fails both the invertibility check and the column-match check. No product \(\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\) can be formed with these matrices.
A Reliable Verification Routine
For a proposed factorization, keep the checks in a fixed order. The order prevents an appealing eigenvalue list from distracting from a missing inverse or a mismatched column. When \(\mathbf{P}^{-1}\) is supplied, multiply back to check it; when it is not supplied, compute or otherwise establish the inverse before using the factorization.
Confirm that \(\mathbf{A}\) and \(\mathbf{P}\) are \(n\times n\), that \(\mathbf{D}\) is \(n\times n\), and that all off-diagonal entries of \(\mathbf{D}\) are zero.
If a candidate inverse is given, verify its products with \(\mathbf{P}\). If no inverse exists, the proposed factorization cannot be valid.
For every column \(\mathbf{p}_j\), calculate \(\mathbf{A}\mathbf{p}_j\) and compare it with \(\lambda_j\mathbf{p}_j\), where \(\lambda_j\) is the \(j\)-th diagonal entry of \(\mathbf{D}\).
Check \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\). If this holds and \(\mathbf{P}\) is invertible, the theorem proves the claimed factorization. As a final direct check, multiply \(\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\) and compare it with \(\mathbf{A}\).
Common Pitfall: Matching Eigenvalues Without Matching Positions
A common incorrect argument is: “The entries \(2\) and \(3\) are eigenvalues of \(\mathbf{A}\), so \(\mathbf{D}=\begin{bmatrix}3&0\\0&2\end{bmatrix}\) must work.” The ordering example shows the error concretely. For \(\mathbf{P}=\mathbf{I}_2\), the first column is \((1,0)^{\top}\), and its image is \((2,0)^{\top}\), not \(3(1,0)^{\top}=(3,0)^{\top}\). The second column maps to \((0,3)^{\top}\), not \(2(0,1)^{\top}=(0,2)^{\top}\). The correction is to pair each column with its own eigenvalue, or to reorder the columns and diagonal entries together.
Another useful safeguard is to distinguish a successful shortcut check from an incomplete one. The equation \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\) alone does not license writing \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\) unless \(\mathbf{P}\) is invertible. Conversely, when invertibility has been verified, the shortcut is not merely suggestive: the theorem proves it is equivalent to the full factorization.
Check Your Understanding
For each proposed diagonalization, identify which checks are needed before accepting the factorization.
- State the equation that is equivalent to \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\) when \(\mathbf{P}\) is invertible.
- If the third column of \(\mathbf{P}\) is \(\mathbf{p}_3\) and the third diagonal entry of \(\mathbf{D}\) is \(\lambda_3\), what equation must be checked?
- Why does \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\) not by itself establish a diagonalization if \(\mathbf{P}\) is singular?
- For \(\mathbf{A}=\begin{bmatrix}4&0\\0&1\end{bmatrix}\), \(\mathbf{P}=\begin{bmatrix}0&1\\1&0\end{bmatrix}\), and \(\mathbf{D}=\begin{bmatrix}1&0\\0&4\end{bmatrix}\), check the ordered column equations.
- Explain how multiplying \(\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\) back and comparing with \(\mathbf{A}\) verifies the proposed factorization.
