Polynomial Evaluation on Eigenvectors
For a normal operator, an orthonormal eigenbasis turns a polynomial calculation into scalar arithmetic. Each eigenvector is acted on by the operator as multiplication by its eigenvalue, so applying a polynomial to the operator acts on that eigenvector by applying the same polynomial to the eigenvalue. This gives a direct way to compute powers, identify eigenspaces, and test whether a polynomial function is invertible.
Throughout, let \(V\) be a nonzero finite-dimensional inner-product space over \(\mathbb{C}\), and let \(T:V\to V\) be normal, meaning \(TT^*=T^*T\). By the Complex Spectral Theorem, \(V\) has an orthonormal basis of eigenvectors of \(T\). For an eigenvalue \(\lambda\), its eigenspace is \(E_\lambda=\{\mathbf{v}\in V:T\mathbf{v}=\lambda\mathbf{v}\}\). The distinct eigenspaces are mutually orthogonal and their direct sum is \(V\).
Write \(\sigma(T)\) for the finite set of distinct eigenvalues of \(T\). For each \(\lambda\in\sigma(T)\), let \(P_\lambda\) be the orthogonal projection onto \(E_\lambda\). The orthogonal decomposition into eigenspaces gives \(\mathbf{v}=\sum_{\lambda\in\sigma(T)}P_\lambda\mathbf{v}\) for every \(\mathbf{v}\in V\). Since \(T\) acts as \(\lambda\) on \(E_\lambda\), it follows that \(T=\sum_{\lambda\in\sigma(T)}\lambda P_\lambda\). Polynomial evaluation therefore gives the corresponding formula for \(p(T)\).
Proof. For any \(\mathbf{v}\in V\), decompose it as \(\mathbf{v}=\sum_{\lambda\in\sigma(T)}\mathbf{v}_\lambda\), where \(\mathbf{v}_\lambda=P_\lambda\mathbf{v}\in E_\lambda\). For every integer \(k\geq 0\), \(T^k\mathbf{v}_\lambda=\lambda^k\mathbf{v}_\lambda\): for \(k=0\) this is the identity action, and each further application of \(T\) multiplies the vector by \(\lambda\). Thus \(p(T)\mathbf{v}_\lambda=p(\lambda)\mathbf{v}_\lambda\). Summing over the eigenspace components proves the displayed formula for \(p(T)\).
Every nonzero vector in \(E_\lambda\) is consequently an eigenvector of \(p(T)\) with eigenvalue \(p(\lambda)\). Conversely, if a nonzero vector \(\mathbf{v}=\sum_\lambda\mathbf{v}_\lambda\) satisfies \(p(T)\mathbf{v}=\mu\mathbf{v}\), then \(\sum_\lambda (p(\lambda)-\mu)\mathbf{v}_\lambda=\mathbf{0}\). The decomposition into the direct sum of eigenspaces is unique, so \((p(\lambda)-\mu)\mathbf{v}_\lambda=\mathbf{0}\) for every \(\lambda\). Hence any nonzero component belongs to an eigenspace with \(p(\lambda)=\mu\). This proves the stated eigenspace formula and shows that no other eigenvalue \(\mu\) can occur. Finally, an orthonormal eigenbasis for \(T\) is also an orthonormal eigenbasis for \(p(T)\), by the component calculation. In that basis \(p(T)\) is diagonal, so it is normal. \(\square\)
Calculating Polynomial Functions and Powers
Worked Example: A Polynomial with Complex Eigenvalues
Let \(V=\mathbb{C}^3\) with its standard inner product, and let \(T\) have matrix \([T]_{\mathcal{B}\leftarrow\mathcal{B}}=\begin{bmatrix}1&0&0\\0&i&0\\0&0&2\end{bmatrix}\) in the standard orthonormal basis \(\mathcal{B}=(\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3)\). Compute \(p(T)\) for \(p(z)=z^2+z-1\), and state its eigenvalues and eigenspaces.
The operator acts diagonally, so evaluate the polynomial at each diagonal entry: \(p(1)=1+1-1=1\), \(p(i)=i^2+i-1=-1+i-1=-2+i\), and \(p(2)=4+2-1=5\). Therefore
The eigenvalues are \(1,-2+i,5\), with eigenspaces \(\operatorname{span}\{\mathbf{e}_1\}\), \(\operatorname{span}\{\mathbf{e}_2\}\), and \(\operatorname{span}\{\mathbf{e}_3\}\), respectively. To check the computation directly, \(T\mathbf{e}_1=\mathbf{e}_1\), so \(p(T)\mathbf{e}_1=(1+1-1)\mathbf{e}_1=\mathbf{e}_1\). Also \(T\mathbf{e}_2=i\mathbf{e}_2\), giving \(p(T)\mathbf{e}_2=(-2+i)\mathbf{e}_2\); and \(T\mathbf{e}_3=2\mathbf{e}_3\), giving \(p(T)\mathbf{e}_3=5\mathbf{e}_3\). These checks reproduce each diagonal entry.
Powers are a special case: for \(p(z)=z^k\), \(p(T)=T^k\), and the eigenvalue \(\lambda\) becomes \(\lambda^k\). Distinct eigenvalues can, however, become the same value after applying a polynomial. When that happens, the eigenspace of \(p(T)\) is the direct sum of the original eigenspaces that map to that value; it need not equal just one of them.
Worked Example: A Power Merges Eigenspaces
Let \(T:\mathbb{C}^4\to\mathbb{C}^4\) have matrix \([T]_{\mathcal{B}\leftarrow\mathcal{B}}=\begin{bmatrix}1&0&0&0\\0&-1&0&0\\0&0&1&0\\0&0&0&2\end{bmatrix}\) in the standard orthonormal basis. Find the eigenvalues and eigenspaces of \(T^2\) and \(T^4\).
For \(T^2\), apply \(z\mapsto z^2\) to the entries \(1,-1,1,2\). The results are \(1,1,1,4\), so \([T^2]_{\mathcal{B}\leftarrow\mathcal{B}}=\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&4\end{bmatrix}\). Its eigenvalue \(1\) has eigenspace \(\operatorname{span}\{\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3\}\), and eigenvalue \(4\) has eigenspace \(\operatorname{span}\{\mathbf{e}_4\}\).
For \(T^4\), the scalar values are \(1^4=1\), \((-1)^4=1\), \(1^4=1\), and \(2^4=16\). Thus \([T^4]_{\mathcal{B}\leftarrow\mathcal{B}}=\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&16\end{bmatrix}\). The eigenvalue \(1\) again has eigenspace \(\operatorname{span}\{\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3\}\), and \(16\) has eigenspace \(\operatorname{span}\{\mathbf{e}_4\}\).
Check: Applying \(T^2\) twice multiplies \(\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3\) by \(1\) and \(\mathbf{e}_4\) by \(4^2=16\), agreeing with the displayed matrix for \(T^4\). In particular, the two distinct eigenvalues \(1\) and \(-1\) of \(T\) both map to \(1\) under the square, so both original eigenspaces belong to the eigenspace for \(1\) of \(T^2\).
When a Polynomial Function Is Invertible
The eigenvalue formula also gives an exact invertibility test. The operator \(p(T)\) is invertible precisely when none of its eigenvalues is zero, which means \(p(\lambda)\ne 0\) for every \(\lambda\in\sigma(T)\). When this condition holds, its inverse acts on \(E_\lambda\) by multiplication by \(1/p(\lambda)\). The finite eigenvalue set also lets us express this inverse as a polynomial in \(T\).
Proof. If \(p(\lambda)=0\) for some eigenvalue \(\lambda\), choose a nonzero \(\mathbf{v}\in E_\lambda\). Then \(p(T)\mathbf{v}=p(\lambda)\mathbf{v}=\mathbf{0}\), so \(p(T)\) has a nonzero vector in its kernel and cannot be invertible. Conversely, suppose \(p(\lambda)\ne0\) for every eigenvalue. Define \(S=\sum_{\lambda\in\sigma(T)}(1/p(\lambda))P_\lambda\). Since \(P_\lambda P_\nu=\mathbf{O}\) for distinct eigenspaces and \(P_\lambda^2=P_\lambda\), the formula for \(p(T)\) gives \(p(T)S=\sum_\lambda P_\lambda=\operatorname{id}_V\) and \(Sp(T)=\sum_\lambda P_\lambda=\operatorname{id}_V\). Thus \(S=p(T)^{-1}\).
To obtain a polynomial, define $$ q(z)=\sum_{\lambda\in\sigma(T)} \frac{1}{p(\lambda)} \prod_{\substack{\nu\in\sigma(T)\\\nu\ne\lambda}} \frac{z-\nu}{\lambda-\nu}. $$ All denominators are nonzero because the sum uses distinct eigenvalues. At any \(\lambda\in\sigma(T)\), its own product equals \(1\), while each other term has a factor \(z-\lambda\) and equals \(0\). Hence \(q(\lambda)=1/p(\lambda)\). Applying the polynomial formula already proved gives \(q(T)=\sum_\lambda q(\lambda)P_\lambda=\sum_\lambda(1/p(\lambda))P_\lambda=p(T)^{-1}\). The product over no factors, if there is only one distinct eigenvalue, is defined to be \(1\), so the argument covers that case too. \(\square\)
Worked Example: Inverting a Polynomial Function
Let \(T:\mathbb{C}^2\to\mathbb{C}^2\) have matrix \([T]_{\mathcal{B}\leftarrow\mathcal{B}}=\begin{bmatrix}1&0\\0&2\end{bmatrix}\) in the standard orthonormal basis, and take \(p(z)=z^2+1\). Compute \(p(T)^{-1}\) and express it as a polynomial in \(T\).
The eigenvalues of \(T\) are \(1\) and \(2\), and \(p(1)=2\), \(p(2)=5\). Both are nonzero, so the inverse exists. Direct evaluation gives \([p(T)]_{\mathcal{B}\leftarrow\mathcal{B}}=\begin{bmatrix}2&0\\0&5\end{bmatrix}\) and \([p(T)^{-1}]_{\mathcal{B}\leftarrow\mathcal{B}}=\begin{bmatrix}\tfrac12&0\\0&\tfrac15\end{bmatrix}\).
To find a linear polynomial \(q(z)=az+b\) with \(q(1)=\tfrac12\) and \(q(2)=\tfrac15\), solve \(a+b=\tfrac12\) and \(2a+b=\tfrac15\). Subtraction gives \(a=-\tfrac{3}{10}\), and then \(b=\tfrac12+\tfrac{3}{10}=\tfrac45\). Therefore \(q(T)=-\tfrac{3}{10}T+\tfrac45\operatorname{id}_V\), whose matrix is
Check: Multiplication gives \(\begin{bmatrix}2&0\\0&5\end{bmatrix}\begin{bmatrix}\tfrac12&0\\0&\tfrac15\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}\). Thus the polynomial expression and the computed inverse agree.
Polynomial Arithmetic and a Common Mistake
Polynomial evaluation respects addition and multiplication: for polynomials \(p\) and \(r\), \((p+r)(T)=p(T)+r(T)\) and \((pr)(T)=p(T)r(T)\). For the product, write \(p(z)=\sum_j a_jz^j\) and \(r(z)=\sum_k b_kz^k\). Then \(p(T)r(T)=\sum_j\sum_k a_jb_kT^{j+k}\), because powers of the same operator multiply by \(T^jT^k=T^{j+k}\). Grouping terms by exponent gives exactly the coefficients of \((pr)(z)\), proving the rule. This concerns polynomials in the same operator; no assumption about unrelated operators is needed.
Common mistake: squaring a sum without its cross term. A tempting but incorrect calculation is \((T+\operatorname{id}_V)^2=T^2+\operatorname{id}_V\). In fact, distributing the product gives \((T+\operatorname{id}_V)^2=T^2+T+T+\operatorname{id}_V=T^2+2T+\operatorname{id}_V\). For \(T=\begin{bmatrix}1&0\\0&2\end{bmatrix}\), the incorrect expression gives \(\begin{bmatrix}2&0\\0&5\end{bmatrix}\), while the correct expression gives \(\begin{bmatrix}4&0\\0&9\end{bmatrix}\). Directly, \(T+\operatorname{id}_V=\begin{bmatrix}2&0\\0&3\end{bmatrix}\), whose square is indeed \(\begin{bmatrix}4&0\\0&9\end{bmatrix}\). Equivalently, the scalar polynomial is \((z+1)^2=z^2+2z+1\), not \(z^2+1\).
The eigenvalue viewpoint makes the main uses concise: evaluate \(p\) on the spectrum to find the eigenvalues of \(p(T)\); collect all original eigenspaces that produce the same value to find its eigenspace; and check whether any value is zero to decide invertibility. These rules apply because normality in finite dimension supplies an orthonormal eigenbasis, so every vector can be analyzed by its components in the eigenspaces.
Check Your Understanding
Use an orthonormal eigenbasis and verify each conclusion from the eigenvalues.
- If the eigenvalues of a normal operator \(T\) are \(1\), \(-2\), and \(i\), what are the possible eigenvalues of \(T^3\)?
- Suppose \(T\) has eigenspaces \(E_1\) and \(E_{-1}\). Describe the eigenspace of \(T^2\) for eigenvalue \(1\).
- For \(p(z)=z^2-4\), state the condition on \(\sigma(T)\) that makes \(p(T)\) invertible.
- Let \(T\) have eigenvalues \(2\) and \(3\), with corresponding eigenspaces \(E_2\) and \(E_3\). Find the eigenspaces of \(T^2-T\).
- For \(p(z)=z+1\), expand \(p(T)^2\) as a polynomial in \(T\), including every term.
