Consider the linear transformation T: R4 → R4, given by T(x,y,z,w) = (2x+z,−y+z,−5z,3w), with its matrix A on the canonical basis of R4. Choose an option: (a) The vector v1 = (1,0,0,0) satisfies A(v1) = 2v1, so it is an eigenvector associated with the eigenvalue λ1 = 2. Beyond that, there are vectors v2,v3,v4 with A(v2)= −v2, A(v3) = −5v3 and A(v4) = −4v4. (b) The vector v1 = (1,0,0,0) satisfies A(v1) = 3v1, so it is an eigenvector associated with the eigenvalue λ1 = 3. Beyond that, there are vectors v2,v3,v4 with A(v2) = v2, A(v3) = −5v3 and A(v4) = −4v4. (c) The vector v1 = (1,0,0,0) satisfies A(v1) = −2v1, so it is an eigenvector associated with the eigenvalue λ1 = −2. Beyond that, there are vectors v2,v3,v4 with A(v2) = v2, A(v3) = 5v3 and A(v4) = −4v4. (d) The vector v1 = (1,0,0,0) satisfies A(v1) = 2v1, so it is an eigenvector associated with the eigenvalue λ1 = 2. Beyond that, there are vectors v2,v3,v4 with A(v2) = −v2, A(v3) = −5v3 and A(v4) = −4v4. (e) The vector v1 = (1,0,0,0) satisfies A(v1) = 3v1, so it is an eigenvector associated with the eigenvalue λ1 = 3. Beyond that, there are vectors v2,v3,v4 with A(v2) = −2v2, A(v3) = −5v3 and A(v4) = −4v4.
Consider the linear transformation T: R4 → R4, given by T(x,y,z,w) = (2x+z,−y+z,−5z,3w), with its matrix A on the canonical basis of R4.
Choose an option:
(a) The vector v1 = (1,0,0,0) satisfies A(v1) = 2v1, so it is an eigenvector associated with the eigenvalue λ1 = 2. Beyond that, there are
(b) The vector v1 = (1,0,0,0) satisfies A(v1) = 3v1, so it is an eigenvector associated with the eigenvalue λ1 = 3. Beyond that, there are vectors v2,v3,v4 with A(v2) = v2, A(v3) = −5v3 and A(v4) = −4v4.
(c) The vector v1 = (1,0,0,0) satisfies A(v1) = −2v1, so it is an eigenvector associated with the eigenvalue λ1 = −2. Beyond that, there are vectors v2,v3,v4 with A(v2) = v2, A(v3) = 5v3 and A(v4) = −4v4.
(d) The vector v1 = (1,0,0,0) satisfies A(v1) = 2v1, so it is an eigenvector associated with the eigenvalue λ1 = 2. Beyond that, there are vectors v2,v3,v4 with A(v2) = −v2, A(v3) = −5v3 and A(v4) = −4v4.
(e) The vector v1 = (1,0,0,0) satisfies A(v1) = 3v1, so it is an eigenvector associated with the eigenvalue λ1 = 3. Beyond that, there are vectors v2,v3,v4 with A(v2) = −2v2, A(v3) = −5v3 and A(v4) = −4v4.
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