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Unformatted text preview: Solutions to problems from section 5.3
subspaces hhhh of R( 2. Let D = {d1 , d2 } and B = {b1 , b2 } be bases for (((((( vector spaces hh h
m V and W , respectively. Let T : V → W be a linear transformation with the property that T (d1 ) = 2b1 − 3b2 , T (d2 ) = −4b1 + 5b2 . −4 5 Find the matrix for T relative to D and B . 2 Solution: Since [T (d1 )]B = and [T (d2 )]B = −3 2 −4 relative to D and B is . −3 5
subspace of R h (
m we see that the matrix for T hh space vectorhh 4. Let B = {b1 , b2 , b3 } be a basis of a (((((h V and T : V → R2 be a linear transformation with the property that T (x1 b1 + x2 b2 + x3 b3 ) = 2x1 − 4x2 + 5x3 −x2 + 3x3 . Find the matrix for T relative to B and the standard matrix for R2 . 2 −4 Solution: Since [T (b1 )]E = , [T (b2 )]E = , and [T (b3 )]E = 0 −1 2 −4 5 . the matrix for T relative to B and E is 0 −1 3
subspace of R h (
m 5 3 we see that hh space vectorhh 8. Let B = {b1 , b2 , b3 } be a basis of a (((((h V . Find T (3b1 − 4b2 ) when T is a linear transformation from V to V whose matrix relative to B is 0 −6 1 [T ]B = 0 5 −1 . 1 −2 7 Solution: 0 −6 1 3 24 5 −1 −4 = −20 . [T (3b1 − 4b2 )]B = 0 1 −2 7 0 11 3 2 −1 1 −1 4 . −2 3 Thus T (3b1 − 4b2 ) = 24b1 − 20b2 + 11b3 . 12. Find the B matrix for the transformation x → Ax when B = 3 2 standard matrix A. Notice Solution: Let b1 = T (b1 ) = T (b2 ) = Thus [T ]B = −1 4 −2 3 −1 4 −2 3 . and b2 = −1 1 5 0 5 5 , and A = and let T denote the transformation with 3 2 −1 1 = = b1 − 2b2 = 2b1 + 1b2 ⇒ [T (b1 ]B = ⇒ [T (b2 ]B = 1 −2 2 1 , = . 12 −2 1 14. Deﬁne T : R2 → R2 by T (x) = Ax where A = property that [T ]B is diagonal. 5 −3 . Find a basis B for R2 with the −7 1 Solution: We know A = P [T ]B P −1 where the columns of P are the elements of B , so by theorem 5 in section 5.3 we know [T ]B is diagonal if and only if B is a basis of eigenvectors of A. So we need to ﬁnd an eigenvector basis of R2 . First we ﬁnd the eigenvalues: χA (λ) = det 5 − λ −3 −7 1 − λ = (5 − λ)(1 − λ) − 21 = λ2 − 6λ − 16 = (λ − 8)(λ + 2), so the eigenvalues of A are 8 and −2 (both with multiplicity 1). Since A − 8I2 = −3 −3 −7 −7 ∼ 11 00 we see that x ∈ Nul(A − 8I2 ) if and only if x= −1 1 x1 x2 = − x2 x2 = x2 −1 1 so is a basis for the 8eigenspace of A. Since 7 −3 −7 3 1 −3/7 0 0 A + 2I2 = ∼ we see that x ∈ Nul(A + 2I2 ) if and only if x= 3 7 x1 x2 = 3/7x2 x2 = x2 3/7 1 so is a basis for the −2eigenspace of A. Set −1 1 3 7 B= , . 16. Deﬁne T : R2 → R2 by T (x) = Ax where A = property that [T ]B is diagonal. 2 −6 . Find a basis B for R2 with the −1 3 Solution: Again, we need to ﬁnd an eigenvector basis of R2 . First we ﬁnd the eigenvalues: χA (λ) = det 2 − λ −6 −1 3 − λ = (2 − λ)(3 − λ) − 6 = λ2 − 5λ = λ(λ − 5), so the eigenvalues of A are 0 and 5 (both with multiplicity 1). Since A∼ 1 −3 0 0 we see that x ∈ Nul(A) if and only if x= 3 1 x1 x2 = 3x2 x2 = x2 3 1 so is a basis for the 0eigenspace of A. Since −3 −6 −1 −2 12 00 A − 5I2 = ∼ we see that x ∈ Nul(A − 5I2 ) if and only if x= −2 1 x1 x2 = −2x2 x2 = x2 −2 1 so is a basis for the 5eigenspace of A. Set 3 1 −2 1 B= , . 25. The trace of a square matrix A is the sum of the diagonal entries in A and is denoted trA. It can be veriﬁed that tr(F G) = tr(GF ) for any two n × n matrices F and G. Show that if A and B are similar then trA = trB . Solution: If A and B are similar matrices then there exists an invertible matrix P with A = P BP −1 . Thus trA = tr(P BP −1 ) = tr(P (BP −1 )) = tr((BP −1 )P ) = tr(BP −1 P ) = tr(BI ) = trB. 26. It can be shown that the trace of a matrix A is the sum of the eigenvalues of A. Verify this statement for the case when A is diagonalizable. Solution: Suppose the eigenvalues of A are λ1. . . , λn . If A is diagonalizable then A is , λ1 .. similar to a diagonal matrix D = . By the previous exercise we know . λn trA = trD = λ1 + · · · + λn .
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 Fall '08
 AXENOVICH
 Vector Space

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