Eigenvalues and Eigenvectors Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Eigenvalues and Eigenvectors Consider the equation 𝐴 𝑥 = 𝜆 𝑥 , where A is an nxn matrix. We call 𝑥 (must be non-zero) an eigenvector of A if this equation can be solved for some value of 𝜆. We call 𝜆 an eigenvalue of matrix A, and we call 𝑥 an eigenvector corresponding to 𝜆. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Eigenvalues and Eigenvectors Consider the equation 𝐴 𝑥 = 𝜆 𝑥 , where A is an nxn matrix. We call 𝑥 (must be non-zero) an eigenvector of A if this equation can be solved for some value of 𝜆. We call 𝜆 an eigenvalue of matrix A, and we call 𝑥 an eigenvector corresponding to 𝜆. We can rearrange the equation to see how to solve for 𝜆 and 𝑥 : 𝜆 is an eigenvalue of A if, and only if the equation 𝐴−𝜆𝐼 𝑥 = 0 has a nontrivial solution. So the set of solutions is just the nullspace of (𝐴−𝜆𝐼). This is a subspace of ℝn and we call it the eigenspace corresponding to the eigenvalue 𝜆. nxn identity matrix Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Eigenvalues and Eigenvectors Example: Consider the matrix 𝐴= 3 2 3 8 . An eigenvalue of this matrix is 𝜆=2. Find the associated eigenvectors. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Eigenvalues and Eigenvectors Example: Consider the matrix 𝐴= 3 2 3 8 . An eigenvalue of this matrix is 𝜆=2. Find the associated eigenvectors. We need to find the nullspace of 𝐴−𝜆𝐼 : 𝐴−2𝐼= 3 2 3 8 − 2 0 0 2 = 1 2 3 6 Nullspace will be vectors satisfying 𝑥 1 + 2𝑥 2 =0 𝑥 = −2 1 is an eigenvector. The eigenspace is the 1-dimensional subspace of ℝ2 consisting of all multiples of −2 1 . Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Eigenvalues and Eigenvectors Example: Consider the matrix 𝐴= 3 2 3 8 . An eigenvalue of this matrix is 𝜆=2. Find the associated eigenvectors. We need to find the nullspace of 𝐴−𝜆𝐼 : 𝐴−2𝐼= 3 2 3 8 − 2 0 0 2 = 1 2 3 6 Nullspace will be vectors satisfying 𝑥 1 + 2𝑥 2 =0 𝑥 = −2 1 is an eigenvector. The eigenspace is the 1-dimensional subspace of ℝ2 consisting of all multiples of −2 1 . We can check that the eigenvector equation 𝐴 𝑥 = 𝜆 𝑥 is satisfied: 3 2 3 8 −2 1 = −4 2 =2 −2 1 √ Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Eigenvalues and Eigenvectors We know how to find eigenvectors, but only if we know what eigenvalue to use. How do we find eigenvalues? Here is the equation we need to solve, and we want to find non-trivial solutions: 𝐴−𝜆𝐼 𝑥 = 0 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Eigenvalues and Eigenvectors We know how to find eigenvectors, but only if we know what eigenvalue to use. How do we find eigenvalues? Here is the equation we need to solve, and we want to find non-trivial solutions: 𝐴−𝜆𝐼 𝑥 = 0 The only way to get non-trivial solutions is if the determinant is 0. This gives us an equation we can solve for 𝜆. A scalar 𝜆 is an eigenvalue of an nxn matrix A if and only if 𝜆 satisfies the characteristic equation det 𝐴−𝜆𝐼 =0. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Eigenvalues and Eigenvectors We know how to find eigenvectors, but only if we know what eigenvalue to use. How do we find eigenvalues? Here is the equation we need to solve, and we want to find non-trivial solutions: 𝐴−𝜆𝐼 𝑥 = 0 The only way to get non-trivial solutions is if the determinant is 0. This gives us an equation we can solve for 𝜆. A scalar 𝜆 is an eigenvalue of an nxn matrix A if and only if 𝜆 satisfies the characteristic equation det 𝐴−𝜆𝐼 =0. Solving the characteristic equation will give all the eigenvalues of the matrix. They could be real (possibly repeated) or complex. We will look at several examples. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Eigenvalues and Eigenvectors Example 1: Consider the matrix 𝐴= 3 2 3 8 . Find all the eigenvalues and their associated eigenspaces. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Eigenvalues and Eigenvectors Example 1: Consider the matrix 𝐴= 3 2 3 8 . Find all the eigenvalues and their associated eigenspaces. First we solve the characteristic equation to find the e-values: 3−𝜆 2 3 8−𝜆 =0 → 3−𝜆 8−𝜆 −6=0 → 𝜆 2 −11𝜆+18=0 𝜆−2 𝜆−9 =0 → 𝜆=2 𝑜𝑟 𝜆=9 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Eigenvalues and Eigenvectors Example 1: Consider the matrix 𝐴= 3 2 3 8 . Find all the eigenvalues and their associated eigenspaces. First we solve the characteristic equation to find the e-values: 3−𝜆 2 3 8−𝜆 =0 → 3−𝜆 8−𝜆 −6=0 → 𝜆 2 −11𝜆+18=0 𝜆−2 𝜆−9 =0 → 𝜆=2 𝑜𝑟 𝜆=9 Here we have 2 distinct, real eigenvalues. We expect to get a 1-dimensional eigenspace for each of them. We already found the e-vectors for 𝜆=2. For 𝜆=9 we need to find the nullspace of (A-9I): −6 2 3 −1 →3 𝑥 1 − 𝑥 2 =0 → 𝑥 = 1 3 . Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Eigenvalues and Eigenvectors Example 1: Consider the matrix 𝐴= 3 2 3 8 . Find all the eigenvalues and their associated eigenspaces. Here are our results: Eigenvalue Eigenspace 𝜆 1 =2 𝑠𝑝𝑎𝑛 −2 1 𝜆 2 =9 𝑠𝑝𝑎𝑛 1 3 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Eigenvalues and Eigenvectors Example 1: Consider the matrix 𝐴= 3 2 3 8 . Find all the eigenvalues and their associated eigenspaces. Here are our results: Eigenvalue Eigenspace 𝜆 1 =2 𝑠𝑝𝑎𝑛 −2 1 𝜆 2 =9 𝑠𝑝𝑎𝑛 1 3 We can give this a geometrical interpretation. The matrix A represents a linear transformation from ℝ2→ ℝ2, and the eigenspaces are preserved under that transformation. i.e. any vector that is in an eigenspace is mapped to another vector in that eigenspace (scaled by the eigenvalue). Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Eigenvalues and Eigenvectors Example 1: Consider the matrix 𝐴= 3 2 3 8 . Find all the eigenvalues and their associated eigenspaces. Here are our results: Eigenvalue Eigenspace 𝜆 1 =2 𝑠𝑝𝑎𝑛 −2 1 𝜆 2 =9 𝑠𝑝𝑎𝑛 1 3 This vector is 9 times as long e-vector for λ=9 e-vector for λ=2 This vector is twice as long Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Eigenvalues and Eigenvectors Example 2: Consider the matrix 𝐴= 2 2 1 1 3 1 1 2 2 . Find all the eigenvalues and their associated eigenspaces. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Eigenvalues and Eigenvectors Example 2: Consider the matrix 𝐴= 2 2 1 1 3 1 1 2 2 . Find all the eigenvalues and their associated eigenspaces. First we find the characteristic equation: det 2−𝜆 2 1 1 3−𝜆 1 1 2 2−𝜆 =0 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Eigenvalues and Eigenvectors Example 2: Consider the matrix 𝐴= 2 2 1 1 3 1 1 2 2 . Find all the eigenvalues and their associated eigenspaces. First we find the characteristic equation: det 2−𝜆 2 1 1 3−𝜆 1 1 2 2−𝜆 =0 Cofactor expansion along the first column: 2−𝜆 3−𝜆 1 2 2−𝜆 − 1 2 1 2 2−𝜆 + 1 2 1 3−𝜆 1 =0 2−𝜆 3−𝜆 2−𝜆 −2 − 2 2−𝜆 −2 + 2− 3−𝜆 =0 2−𝜆 𝜆 2 −5𝜆+4 +3𝜆−3=0 2−𝜆 𝜆−1)(𝜆−4 +3(𝜆−1)=0 𝜆−1 𝜆−1 𝜆−5 =0 𝜆=5 𝑜𝑟 𝜆=1 𝑟𝑒𝑝𝑒𝑎𝑡𝑒𝑑 We say the eigenvalue λ=1 has multiplicity 2 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Eigenvalues and Eigenvectors Example 2: Consider the matrix 𝐴= 2 2 1 1 3 1 1 2 2 . Find all the eigenvalues and their associated eigenspaces. Find eigenvectors corresponding to each eigenvalue: 𝐹𝑜𝑟 𝜆=5, 𝑓𝑖𝑛𝑑 𝑛𝑢𝑙𝑙𝑠𝑝𝑎𝑐𝑒 𝑜𝑓 2−5 2 1 1 3−5 1 1 2 2−5 −3 2 1 1 −2 1 1 2 −3 → 1 0 −1 0 1 −1 0 0 0 → 𝑥 = 1 1 1 𝐹𝑜𝑟 𝜆=1, 𝑓𝑖𝑛𝑑 𝑛𝑢𝑙𝑙𝑠𝑝𝑎𝑐𝑒 𝑜𝑓 2−1 2 1 1 3−1 1 1 2 2−1 1 2 1 1 2 1 1 2 1 → 1 2 1 0 0 0 0 0 0 → 𝑥 = 2 −1 0 𝑜𝑟 𝑥 = 1 0 −1 Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Eigenvalues and Eigenvectors Example 2: Consider the matrix 𝐴= 2 2 1 1 3 1 1 2 2 . Find all the eigenvalues and their associated eigenspaces. Here are our results: Eigenvalue Eigenspace 𝜆=1 (multiplicity 2) 𝑠𝑝𝑎𝑛 𝑥 1 = 2 −1 0 , 𝑥 2 = 1 0 −1 𝜆=5 𝑠𝑝𝑎𝑛 𝑥 3 = 1 1 1 This time we have a repeated e-value, and we get a 2-dimensional eigenspace (won’t always happen). So any vector that starts in the plane formed by 𝑥 1 and 𝑥 2 will stay in that plane. In fact, since the e-value is 1, that vector won’t change at all. The other e-value has multiplicity 1, so we get a 1-dimensional eigenspace (a line) just as in the previous example. Any vector starting on that line will stay the same direction, but be transformed to a vector 5 times longer. The diagram on the next slide shows these transformations. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Eigenvalues and Eigenvectors Example 2: Consider the matrix 𝐴= 2 2 1 1 3 1 1 2 2 . Find all the eigenvalues and their associated eigenspaces. 5∙ 𝑥 3 𝑥 3 𝑥 2 1∙ 𝑥 2 1∙ 𝑥 1 𝑥 1 Original e-vectors (zoomed in so you can see them better) Transformed e-vectors Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB