A Complete Diagonalization Check
Diagonalizing a matrix is not just a matter of finding some eigenvalues. A complete solution must establish that there are enough independent eigenvectors, place them in the columns of an invertible matrix, match their order with the diagonal entries, and verify the resulting factorization. Once that work is done, powers of the matrix can be computed by taking powers of the diagonal entries.
All matrices and vectors in this tutorial are real. An eigenvector of a square matrix \(\mathbf{A}\in\mathbb{R}^{n\times n}\) is a nonzero vector \(\mathbf{v}\) for which \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\) for some scalar \(\lambda\); the scalar \(\lambda\) is its eigenvalue. The eigenspace \(E_{\lambda}\) consists of the zero vector and all eigenvectors with eigenvalue \(\lambda\), so \(E_{\lambda}=\operatorname{Nul}(\mathbf{A}-\lambda\mathbf{I})\). The algebraic multiplicity of \(\lambda\) is its multiplicity as a root of \(p_{\mathbf{A}}(\lambda)=\det(\mathbf{A}-\lambda\mathbf{I})\), while its geometric multiplicity is \(\dim E_{\lambda}\).
Proof. Suppose first that \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\), where \(\mathbf{P}\) is invertible and \(\mathbf{D}\) is diagonal with entries \(\lambda_1,\ldots,\lambda_n\). Multiplying on the right by \(\mathbf{P}\) gives \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\). If the columns of \(\mathbf{P}\) are \(\mathbf{p}_1,\ldots,\mathbf{p}_n\), this equation says \(\mathbf{A}\mathbf{p}_i=\lambda_i\mathbf{p}_i\) for each \(i\). The columns are nonzero and linearly independent because \(\mathbf{P}\) is invertible, so they are \(n\) independent eigenvectors. Conversely, suppose \(\mathbf{A}\) has \(n\) linearly independent eigenvectors \(\mathbf{p}_1,\ldots,\mathbf{p}_n\), with respective eigenvalues \(\lambda_1,\ldots,\lambda_n\). Let \(\mathbf{P}\) have those vectors as columns and let \(\mathbf{D}\) have diagonal entries \(\lambda_1,\ldots,\lambda_n\) in the same order. Independence makes \(\mathbf{P}\) invertible, and the eigenvector equations give \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\). Multiplying on the right by \(\mathbf{P}^{-1}\) proves \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\). \(\square\)
This criterion explains the role of eigenspace dimensions. For a given eigenvalue, independent eigenvectors can be chosen only up to the dimension of its eigenspace. A repeated root of the characteristic polynomial does not automatically supply that many independent eigenvectors; the eigenspaces must be checked.
Repeated Eigenvalue, Sufficient Eigenvectors
Worked Example: Diagonalizability with a Repeated Eigenvalue
Test the diagonalizability of the matrix below. If it is diagonalizable, construct \(\mathbf{P}\) and \(\mathbf{D}\), verify the factorization, and use it to find \(\mathbf{A}^4\).
The characteristic polynomial is
Thus the eigenvalues are \(1\), with algebraic multiplicity \(2\), and \(3\), with algebraic multiplicity \(1\). To find \(E_1\), solve \((\mathbf{A}-\mathbf{I})\mathbf{x}=\mathbf{0}\). The equations are \(x-y=0\) and \(z=0\), so \(x=y\), while \(x\) is free. Therefore \(E_1=\operatorname{span}\{(1,1,0),(0,0,1)\}\), which has dimension \(2\). For \(\lambda=3\), the equations from \((\mathbf{A}-3\mathbf{I})\mathbf{x}=\mathbf{0}\) are \(x+y=0\) and \(z=0\). Hence \(E_3=\operatorname{span}\{(1,-1,0)\}\), with dimension \(1\). Together these eigenspaces supply three independent eigenvectors, so the eigenbasis criterion proves that \(\mathbf{A}\) is diagonalizable.
Choose the eigenvectors in the order \((1,1,0)\), \((0,0,1)\), \((1,-1,0)\). The first two have eigenvalue \(1\), and the third has eigenvalue \(3\). Thus
Solving for the coordinates in this eigenvector basis gives
For an inverse check, multiplying the matrices gives
Now compute the factorization itself. The product \(\mathbf{P}\mathbf{D}\) scales the third column of \(\mathbf{P}\) by \(3\), leaving its first two columns unchanged:
For a positive integer \(k\), the factorization gives \(\mathbf{A}^k=\mathbf{P}\mathbf{D}^k\mathbf{P}^{-1}\). In this example, \(\mathbf{D}^4=\operatorname{diag}(1,1,81)\), so
Check: Direct multiplication gives \(\mathbf{A}^2=\begin{bmatrix}5&-4&0\\-4&5&0\\0&0&1\end{bmatrix}\). Squaring this result gives \(\mathbf{A}^4=\begin{bmatrix}5&-4&0\\-4&5&0\\0&0&1\end{bmatrix}^2=\begin{bmatrix}41&-40&0\\-40&41&0\\0&0&1\end{bmatrix}\), agreeing with the diagonalization calculation.
Why the Power Formula Works
Proof. For \(k=0\), the right side is \(\mathbf{P}\mathbf{I}\mathbf{P}^{-1}=\mathbf{I}\), which equals \(\mathbf{A}^0\). For \(k=1\), the statement is the assumed factorization. Suppose for some positive integer \(k\) that \(\mathbf{A}^k=\mathbf{P}\mathbf{D}^k\mathbf{P}^{-1}\). Then, using \(\mathbf{P}^{-1}\mathbf{P}=\mathbf{I}\),
Induction proves the formula for every nonnegative integer \(k\). \(\square\)
The middle cancellation is essential: \(\mathbf{P}^{-1}\mathbf{P}=\mathbf{I}\) allows adjacent diagonal powers to combine. If \(\mathbf{D}\) has diagonal entries \(d_1,\ldots,d_n\), then \(\mathbf{D}^k\) has diagonal entries \(d_1^k,\ldots,d_n^k\). This is usually simpler than multiplying \(\mathbf{A}\) by itself repeatedly.
Distinct Eigenvalues and a Full Power Calculation
Worked Example: Verify the Factorization and Find a Fifth Power
Consider the upper-triangular matrix
Its characteristic polynomial is \((2-\lambda)(3-\lambda)(4-\lambda)\), so the eigenvalues are \(2,3,4\). Vectors \(\mathbf{v}_1=(1,0,0)\), \(\mathbf{v}_2=(1,1,0)\), and \(\mathbf{v}_3=(1,1,1)\) satisfy \(\mathbf{B}\mathbf{v}_1=2\mathbf{v}_1\), \(\mathbf{B}\mathbf{v}_2=3\mathbf{v}_2\), and \(\mathbf{B}\mathbf{v}_3=4\mathbf{v}_3\). These vectors are independent: the matrix having them as columns is triangular with diagonal entries \(1,1,1\), so it is invertible. Thus they form an eigenbasis.
Putting the vectors in that order gives
Indeed, direct multiplication gives \(\mathbf{P}\mathbf{P}^{-1}=\mathbf{I}_3\), and
For the fifth power, \(\mathbf{D}^5=\operatorname{diag}(32,243,1024)\). Therefore
Check: One direct multiplication check is to compute \(\mathbf{B}^4=\begin{bmatrix}16&65&175\\0&81&175\\0&0&256\end{bmatrix}\), then multiply by \(\mathbf{B}\). The entries of \(\mathbf{B}^4\mathbf{B}\) are \(32\) in position \((1,1)\), \(16+195=211\) in position \((1,2)\), \(16+65+700=781\) in position \((1,3)\), \(243\) in position \((2,2)\), \(81+700=781\) in position \((2,3)\), and \(1024\) in position \((3,3)\); the other entries are zero. This recovers the displayed fifth power.
When the Eigenspaces Are Too Small
Worked Example: A Repeated Eigenvalue That Prevents Diagonalization
Test whether the following matrix is diagonalizable over \(\mathbb{R}\):
The characteristic polynomial is \((2-\lambda)^2(3-\lambda)\). For \(\lambda=2\), the equation \((\mathbf{C}-2\mathbf{I})\mathbf{x}=\mathbf{0}\) reduces to \(y=0\) and \(z=0\), with \(x\) free. Thus \(E_2=\operatorname{span}\{(1,0,0)\}\) has dimension \(1\), although \(2\) has algebraic multiplicity \(2\). For \(\lambda=3\), the equations are \(-x+y=0\), \(-y=0\), and \(0=0\), giving \(x=y=0\) and \(z\) free. Hence \(E_3=\operatorname{span}\{(0,0,1)\}\) has dimension \(1\).
Every eigenvector belongs to one of these two eigenspaces, and their dimensions provide only two independent eigenvectors in total. A diagonalization of a \(3\times3\) matrix requires three independent eigenvectors, so the eigenbasis criterion shows that \(\mathbf{C}\) is not diagonalizable over \(\mathbb{R}\). The obstruction is not that the matrix lacks eigenvalues; it is that the eigenspaces do not contain enough independent vectors.
Common Pitfall: Counting Roots as Eigenvectors
A repeated root is counted according to its algebraic multiplicity in the characteristic polynomial; that count does not guarantee an equally large eigenspace. For \(\mathbf{C}\), the eigenvalue \(2\) has algebraic multiplicity \(2\), but its eigenspace has dimension \(1\). It would be wrong to claim that \((1,0,0)\) and \((0,1,0)\) are two eigenvectors for \(\lambda=2\): in fact, \(\mathbf{C}(0,1,0)=(1,2,0)\), which is not \(2(0,1,0)=(0,2,0)\). The correction is to solve \((\mathbf{C}-2\mathbf{I})\mathbf{x}=\mathbf{0}\) and count the dimension of its solution space.
When a diagonalization is available, the columns of \(\mathbf{P}\) and diagonal entries of \(\mathbf{D}\) must stay in matching order. If a column is reordered without reordering the corresponding diagonal entry, the equation \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\) will generally fail. Verifying \(\mathbf{P}\mathbf{D}\mathbf{P}^{-1}=\mathbf{A}\) catches this mismatch, while multiplying back a computed power provides an independent arithmetic check.
Check Your Understanding
For each question, support the conclusion with an eigenspace calculation or a matrix multiplication.
- For the matrix in the first worked example, state the algebraic and geometric multiplicities of each eigenvalue.
- For \(\mathbf{B}\), explain why the order of the columns of \(\mathbf{P}\) must agree with the order of the entries in \(\mathbf{D}\).
- Use the diagonalization of \(\mathbf{B}\) to write \(\mathbf{B}^2\), then check it by direct multiplication.
- For \(\mathbf{C}\), find a basis for each eigenspace and explain why the total number of independent eigenvectors is insufficient.
- State the cancellation used in the proof of the power formula and explain why it applies to \(\mathbf{P}^{-1}\mathbf{P}\).
