Question
Find the characteristic polynomial, eigenvalues, and eigenvectors of a 3x3 matrix
Original question: Example For $$A=\begin{bmatrix} 3 & -4 & 2 \\ 1 & -2 & 2 \\ 1 & -5 & 5 \end{bmatrix},$$ find $c_A(\lambda)$, the eigenvalues of $A$, and the eigenbasis for each eigenvalue.
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Key concept: For a 3×3 matrix, the fastest route is often a direct determinant calculation followed by factorisation. Once the eigenvalues appear, the eigenvectors come from a short null-space solve for each one.
Step by step
Let
3 & -4 & 2 \\ 1 & -2 & 2 \\ 1 & -5 & 5 \end{bmatrix}.$$ ### 1) Characteristic polynomial Compute $$c_A(\lambda)=\det(A-\lambda I).$$ So $$A-\lambda I= \begin{bmatrix} 3-\lambda & -4 & 2 \\ 1 & -2-\lambda & 2 \\ 1 & -5 & 5-\lambda \end{bmatrix}.$$ A direct expansion gives $$c_A(\lambda)=-\lambda^3+4\lambda^2-7\lambda+6.$$ Equivalently, the monic form is $$\lambda^3-4\lambda^2+7\lambda-6.$$ Factorising: $$\lambda^3-4\lambda^2+7\lambda-6=(\lambda-2)(\lambda^2-2\lambda+3).$$ ### 2) Eigenvalues So the eigenvalues are $$\lambda_1=2,\qquad \lambda_2=1+i\sqrt{2},\qquad \lambda_3=1-i\sqrt{2}.$$ ### 3) Eigenbasis for $\lambda=2$ Solve $$(A-2I)\mathbf{x}=0.$$ Then $$A-2I= \begin{bmatrix} 1 & -4 & 2 \\ 1 & -4 & 2 \\ 1 & -5 & 3 \end{bmatrix}.$$ From the equations we get $z=y$ and $x=2y$, so an eigenvector is $$\mathbf{v}_1=(2,1,1).$$ Thus an eigenbasis for $\lambda=2$ is $$\{(2,1,1)\}.$$ ### 4) Complex eigenvalues Over $\mathbb{C}$, each complex eigenvalue has a one-dimensional eigenspace. Any nonzero vector solving $$(A-(1\pm i\sqrt2)I)\mathbf{x}=0$$ generates the corresponding eigenbasis. If your course is working over the reals only, then the only real eigenbasis is the one for $\lambda=2$. ### Pitfall alert A frequent mistake is writing the characteristic polynomial with the wrong sign. If you compute $\det(\lambda I-A)$ instead of $\det(A-\lambda I)$, the polynomial changes by a factor of $(-1)^3=-1$. Another common issue is forgetting that nonreal eigenvalues are allowed only when the field is complex. ### Try different conditions If the question asks for eigenvectors only over $\mathbb{R}$, then you stop after $\lambda=2$ because the other two eigenvalues are not real. If the matrix entries change slightly, the determinant step should be repeated from scratch; even one entry can change the factorisation completely. ### Further reading characteristic polynomial, eigenvalues, eigenvectors