From Eigenvalues to a Diagonalization
A diagonalization turns matrix multiplication into a simpler calculation in coordinates adapted to the matrix. For a real \(3\times3\) matrix, the practical task is to find three eigenvalues and matching eigenvectors, then place those eigenvectors in the columns of a matrix. When the eigenvalues are distinct and real, their eigenvectors provide three independent directions, so the resulting matrix is invertible.
All matrices and vectors in this tutorial are over \(\mathbb{R}\). An eigenvalue \(\lambda\in\mathbb{R}\) of \(\mathbf{A}\in\mathbb{R}^{3\times3}\) is a scalar for which there is a nonzero vector \(\mathbf{v}\in\mathbb{R}^3\) satisfying \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\). Such a vector is an eigenvector. Equivalently, \((\mathbf{A}-\lambda\mathbf{I}_3)\mathbf{v}=\mathbf{0}\). The characteristic polynomial here is \(p_{\mathbf{A}}(\lambda)=\det(\mathbf{A}-\lambda\mathbf{I}_3)\).
For a real eigenvalue, the eigenvector calculation is a homogeneous system over \(\mathbb{R}\). The determinant criterion for square matrices says that such a system has a nonzero solution exactly when its coefficient matrix has determinant zero. Thus the real roots of \(p_{\mathbf{A}}\) are precisely the real eigenvalues. A real matrix can also have nonreal characteristic roots, which are relevant when eigenvalues are considered over \(\mathbb{C}\); here we restrict attention to real roots and real eigenvectors.
A Complete Example: Three Eigenvalues and Three Eigenvectors
Worked Example: Diagonalizing a Companion-Form Matrix
Find the eigenvalues and eigenvectors of \(\mathbf{A}=\begin{bmatrix}0&1&0\\0&0&1\\6&-11&6\end{bmatrix}\), and use them to form an eigenvector matrix.
First compute the characteristic polynomial. Expanding along the first row gives
The roots are the three distinct real numbers \(1,2,3\). For any one of these values, the first two equations in \((\mathbf{A}-\lambda\mathbf{I}_3)\mathbf{v}=\mathbf{0}\), with \(\mathbf{v}=(x,y,z)^{\top}\), are \(-\lambda x+y=0\) and \(-\lambda y+z=0\). Hence \(y=\lambda x\) and \(z=\lambda^2x\). Taking \(x=1\) gives the candidate \(\mathbf{v}=(1,\lambda,\lambda^2)^{\top}\). The third equation becomes \(6-11\lambda+(6-\lambda)\lambda^2=0\), which is equivalent to \(-\lambda^3+6\lambda^2-11\lambda+6=0\), the characteristic-root equation.
Substituting the three roots gives \(\mathbf{v}_1=(1,1,1)^{\top}\), \(\mathbf{v}_2=(1,2,4)^{\top}\), and \(\mathbf{v}_3=(1,3,9)^{\top}\). Check each eigenvector directly:
Use these vectors as columns of \(\mathbf{P}\). Row reduction confirms that \(\mathbf{P}\) is invertible:
There is a pivot in every column, so the square matrix \(\mathbf{P}\) is invertible. With \(\mathbf{D}=\begin{bmatrix}1&0&0\\0&2&0\\0&0&3\end{bmatrix}\), the multiplication check is
Since \(\mathbf{P}\) is invertible, multiplying \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\) on the right by \(\mathbf{P}^{-1}\) gives \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\). The eigenvalue and eigenvector checks, the pivot calculation, and the matrix product all agree.
Why Three Distinct Eigenvalues Give Three Independent Vectors
Proof. Suppose \(c_1\mathbf{v}_1+c_2\mathbf{v}_2+c_3\mathbf{v}_3=\mathbf{0}\). Apply \((\mathbf{A}-\lambda_2\mathbf{I}_3)(\mathbf{A}-\lambda_3\mathbf{I}_3)\) to both sides. The terms involving \(\mathbf{v}_2\) and \(\mathbf{v}_3\) vanish because \(\mathbf{A}\mathbf{v}_j=\lambda_j\mathbf{v}_j\). The remaining term is \(c_1(\lambda_1-\lambda_2)(\lambda_1-\lambda_3)\mathbf{v}_1=\mathbf{0}\). Both scalar factors are nonzero because the eigenvalues are distinct, and \(\mathbf{v}_1\ne\mathbf{0}\), so \(c_1=0\). Applying \((\mathbf{A}-\lambda_1\mathbf{I}_3)(\mathbf{A}-\lambda_3\mathbf{I}_3)\) gives \(c_2=0\), and applying \((\mathbf{A}-\lambda_1\mathbf{I}_3)(\mathbf{A}-\lambda_2\mathbf{I}_3)\) gives \(c_3=0\). Thus the vectors are independent. A square matrix has independent columns exactly when it is invertible, so \(\mathbf{P}=\begin{bmatrix}\mathbf{v}_1&\mathbf{v}_2&\mathbf{v}_3\end{bmatrix}\) is invertible. Column by column, \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\), where \(\mathbf{D}\) has diagonal entries \(\lambda_1,\lambda_2,\lambda_3\). Multiplication by \(\mathbf{P}^{-1}\) gives the required diagonalization. \(\square\)
This result explains why an invertibility check is still valuable in a calculation: it confirms that the eigenvectors were assembled correctly. The proof also shows why the distinctness hypothesis matters. If eigenvalues repeat, this argument cannot separate the corresponding coefficients using nonzero factors; the repeated-eigenvalue case requires a different check of how many independent eigenvectors are available.
Two Further Worked Examples
Worked Example: Eigenvectors from a Symmetric Matrix
Find three eigenpairs for \(\mathbf{B}=\begin{bmatrix}3&0&2\\0&3&0\\2&0&3\end{bmatrix}\), and form its eigenvector matrix.
Expanding \(\det(\mathbf{B}-\lambda\mathbf{I}_3)\) along row 2 gives
The distinct real eigenvalues are \(1,3,5\). For \(\lambda=1\), the first and third equations of \((\mathbf{B}-\mathbf{I}_3)\mathbf{v}=\mathbf{0}\) give \(x+z=0\) and \(x+z=0\), while the second gives \(y=0\); choose \(\mathbf{u}_1=(1,0,-1)^{\top}\). For \(\lambda=3\), the equations give \(x=z=0\), with \(y\) free; choose \(\mathbf{u}_2=(0,1,0)^{\top}\). For \(\lambda=5\), they give \(z=x\) and \(y=0\); choose \(\mathbf{u}_3=(1,0,1)^{\top}\).
Direct multiplication checks all three claims: \(\mathbf{B}\mathbf{u}_1=(1,0,-1)^{\top}\), \(\mathbf{B}\mathbf{u}_2=(0,3,0)^{\top}\), and \(\mathbf{B}\mathbf{u}_3=(5,0,5)^{\top}\). Put the eigenvectors in the stated order:
Adding row 1 to row 3 changes the last row of \(\mathbf{P}\) to \((0,0,2)\), leaving pivots in all three columns. Also, the eigenpair checks give
Thus \(\mathbf{P}\) is invertible and \(\mathbf{B}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\).
Worked Example: Distinct Eigenvalues with Irrational Values
Find the eigenvalues and eigenvectors of \(\mathbf{C}=\begin{bmatrix}2&1&0\\1&2&1\\0&1&2\end{bmatrix}\), and verify that the eigenvector matrix is invertible.
Expanding along the first row, with \(s=2-\lambda\), gives
The roots are \(2\), \(2+\sqrt{2}\), and \(2-\sqrt{2}\), which are real and distinct. For \(\lambda=2\), the system \((\mathbf{C}-2\mathbf{I}_3)\mathbf{v}=\mathbf{0}\) gives \(y=0\) and \(x+z=0\); choose \(\mathbf{w}_0=(1,0,-1)^{\top}\). For \(\lambda=2+s\), where \(s=\sqrt{2}\) or \(s=-\sqrt{2}\), choose \(\mathbf{w}_s=(1,s,1)^{\top}\). Indeed,
because \(s^2=2\), so \((2+s)s=2s+2\). In particular, this verifies both eigenvectors \(\mathbf{w}_{\sqrt{2}}\) and \(\mathbf{w}_{-\sqrt{2}}\). For \(\mathbf{w}_0\), direct multiplication gives \(\mathbf{C}\mathbf{w}_0=(2,0,-2)^{\top}=2\mathbf{w}_0\).
Order the columns as \(\mathbf{w}_{\sqrt{2}},\mathbf{w}_0,\mathbf{w}_{-\sqrt{2}}\). The determinant of the resulting matrix is
So this eigenvector matrix is invertible. With \(\mathbf{D}=\operatorname{diag}(2+\sqrt{2},2,2-\sqrt{2})\), the three verified eigenpair equations give \(\mathbf{C}\mathbf{P}=\mathbf{P}\mathbf{D}\), and hence \(\mathbf{C}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\).
Common Pitfalls and a Reliable Workflow
A cofactor sign error can change the roots. For the first matrix, expanding along row 1 gives a contribution \(+6\) from the second entry: the minor is \(\det\begin{bmatrix}0&1\\6&6-\lambda\end{bmatrix}=-6\), and the cofactor sign is negative, so the contribution is \(-1(-6)=+6\). Writing \(-6\) instead would give \(p_{\mathbf{A}}(0)=-6\), even though \(p_{\mathbf{A}}(0)=\det\mathbf{A}=6\). Checking at \(\lambda=0\) catches the error. The correct polynomial is \(-\lambda(\lambda^2-6\lambda+11)+6\).
Do not call a nonreal root a real eigenvalue. The characteristic polynomial can have nonreal roots even when the matrix has real entries. In this tutorial, only real roots are used with real vectors in \(\mathbb{R}^3\). If a calculation produces nonreal roots, interpreting them as eigenvalues requires working over \(\mathbb{C}\) and allowing complex eigenvectors.
A dependable calculation follows a fixed sequence: compute and factor the characteristic polynomial; identify its real roots; solve each corresponding homogeneous system; check each eigenvector by multiplication; place the vectors as columns of \(\mathbf{P}\); and verify invertibility and \(\mathbf{A}\mathbf{P}=\mathbf{P}\mathbf{D}\). Eigenvectors may be scaled by any nonzero real number, but the order of their columns must match the order of the diagonal entries in \(\mathbf{D}\).
Check Your Understanding
Show the determinant or multiplication checks that support each conclusion.
- For \(\mathbf{A}=\begin{bmatrix}0&1&0\\0&0&1\\6&-11&6\end{bmatrix}\), check that the characteristic polynomial has value \(6\) at \(\lambda=0\) as a consistency check, and then verify the whole polynomial by recomputing the determinant expansion.
- For the same matrix, compute \(\mathbf{A}(1,2,4)^{\top}\) and identify the eigenvalue.
- Explain why the three columns in the eigenvector matrix for \(\mathbf{B}\) are placed in the same order as the entries \(1,3,5\) in \(\mathbf{D}\).
- For \(\mathbf{C}\), why does \(s^2=2\) establish the middle coordinate of \(\mathbf{C}(1,s,1)^{\top}=(2+s)(1,s,1)^{\top}\)?
- State which scalar field and vector space are being used when the tutorial calls a root a real eigenvalue.
