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Chapter 6: Problem 25

Let \(A\) be an \(n \times n\) matrix and let \(\lambda\) be an eigenvalue of \(A .\) Show that if \(B\) is any matrix that commutes with \(A,\) then the eigenspace \(N(A-\lambda I)\) is invariant under \(B\)

Short Answer

Expert verified
To show that the eigenspace \(N(A - \lambda I)\) is invariant under B, we computed \((A - \lambda I)(Bx)\) and found it to be equal to 0. Since \((A - \lambda I)(Bx) = 0\), this means Bx also belongs to the eigenspace \(N(A - \lambda I)\), thus proving the eigenspace is invariant under B.

Step by step solution

01

Determine if A and B are commutative matrices

We are given that A and B are commutative matrices, meaning that \(AB = BA\).
02

Explain the eigenspace of A with respect to 位

According to the problem, 位 is an eigenvalue of A. The eigenspace associated with the eigenvalue 位 is defined as \(N(A-\lambda I) = \{ x : Ax = \lambda x \}\), where I is the identity matrix.
03

Show the eigenspace variance under B

To show that the eigenspace N(A-位I) is invariant under B, we need to prove that for any vector x in the eigenspace, Bx remains in the eigenspace. So, our goal is to prove that \(Bx \in N(A-\lambda I)\) for all \(x \in N(A-\lambda I)\).
04

Begin the proof

Let x be a vector in the eigenspace N(A-位I). Then by definition, \((A-\lambda I)x=0\). Furthermore, we know that A and B commute, so we can compute the product \((A-\lambda I)(Bx)\) using the fact that \(AB = BA\).
05

Compute the product (A-位I)(Bx)

Since A and B are commutative, we have: \((A- \lambda I )(Bx) = A(Bx) - \lambda I (Bx) = (AB)x - \lambda (B)x\) Now, substitute AB with BA (using the commutativity property): \((A- \lambda I)(Bx) = (BA)x - \lambda (B)x = B(Ax) - \lambda (Bx)\)
06

Determine if Bx belongs to eigenspace

Now, as x belongs to the eigenspace N(A-位I), then \(Ax = \lambda x\). We can use this in our previous equation: \((A- \lambda I )(Bx) = B(\lambda x) - \lambda (Bx) = \lambda(Bx) - \lambda (Bx) = 0\) Since \((A- \lambda I)(Bx)=0\), it means that Bx also belongs to the eigenspace N(A-位I). Therefore, we have shown that the eigenspace N(A-位I) is invariant under B.

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Most popular questions from this chapter

Let \(A\) be a \(n \times n\) matrix with Schur decomposition \(U T U^{H} .\) Show that if the diagonal entries of \(T\) are all distinct, then there is an upper triangular matrix \(R\) such that \(X=U R\) diagonalizes \(A\)

For each of the following, find a matrix \(B\) such that \(B^{2}=A\) (a) \(A=\left(\begin{array}{rr}2 & 1 \\ -2 & -1\end{array}\right)\) (b) \(A=\left(\begin{array}{rrr}9 & -5 & 3 \\ 0 & 4 & 3 \\ 0 & 0 & 1\end{array}\right)\)

For each of the following functions, determine whether the given stationary point corresponds to a local minimum, local maximum, or saddle point: (a) \(f(x, y)=3 x^{2}-x y+y^{2} \quad(0,0)\) (b) \(f(x, y)=\sin x+y^{3}+3 x y+2 x-3 y \quad(0,-1)\) (c) \(f(x, y)=\frac{1}{3} x^{3}-\frac{1}{3} y^{3}+3 x y+2 x-2 y \quad(1,-1)\) (d) \(f(x, y)=\frac{y}{x^{2}}+\frac{x}{y^{2}}+x y \quad(1,1)\) (e) \(f(x, y, z)=x^{3}+x y z+y^{2}-3 x \quad(1,0,0)\) (f) \(f(x, y, z)=-\frac{1}{4}\left(x^{-4}+y^{-4}+z^{-4}\right)+y z-x-\) \(2 y-2 z \quad(1,1,1)\)

Let \\[ A=\left(\begin{array}{rr} 1 & -\frac{1}{2} \\ -\frac{1}{2} & 1 \end{array}\right) \quad \text { and } \quad B=\left(\begin{array}{rr} 1 & -1 \\ 0 & 1 \end{array}\right) \\] (a) Show that \(A\) is positive definite and that \(\mathbf{x}^{T} A \mathbf{x}=\mathbf{x}^{T} B \mathbf{x}\) for all \(\mathbf{x} \in \mathbb{R}^{2}\) (b) Show that \(B\) is positive definite, but \(B^{2}\) is not positive definite.

Show that the diagonal entries of a Hermitian matrix must be real.
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