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Chapter 6: Problem 61

Show that all matrices similar to an invertible matrix are invertible. More generally, show that similar matrices have the same rank.

Short Answer

Expert verified
All matrices similar to an invertible matrix are invertible: Let A be an invertible matrix, and B be a matrix similar to A. By definition, there exists an invertible matrix P such that B = \(P^{-1}AP\). Since A is invertible and P is invertible, we can apply the associative property of matrix multiplication to obtain \((PA)P^{-1} = PB = (PA)P^{-1}\). Thus, we find that B is equal to A and therefore invertible. Similar matrices have the same rank: For matrices A and B such that B = \(P^{-1}AP\), we can apply a transformation to a vector x in the column space of A to obtain \(P^{-1}APx = P^{-1}y\). This implies that the column space of B is related to the column space of A via the invertible matrix \(P^{-1}\), which is an injective linear transformation. Since \(P^{-1}\) is invertible, it maps basis vectors in the column space of A to linearly independent vectors in the column space of B, retaining the same dimension for both column spaces. Therefore, similar matrices have the same rank.

Step by step solution

01

Recall the definition of invertibility

An nxn matrix A is invertible if there exists another matrix A^(-1) such that AA^(-1) = A^(-1)A = I, where I is the identity matrix. In other words, an invertible matrix has a unique inverse matrix that, when multiplied with the original matrix, results in the identity matrix.
02

Let matrix B be similar to an invertible matrix A

Assume that matrix A is invertible, and matrix B is similar to matrix A. According to the definition of similar matrices, there exists an invertible matrix P such that B = P^(-1)AP.
03

Prove that matrix B is invertible

Consider the product (PA)P^(-1). Using the associative property of matrix multiplication, we can rewrite this expression as P(AP^(-1)). Since B = P^(-1)AP, the product (PA)P^(-1) is equal to PB. We can now multiply both sides by the inverse of P and use the properties of invertibility: \(PB = (PA)P^{-1}\) \(P^{-1}PB = P^{-1}(PA)P^{-1}\) \(B = A(P^{-1}P)\) \(B = AI\) \(B = A\) We find that the original matrix B is equal to A, meaning that B is invertible since A is invertible.
04

Show that similar matrices have the same rank

Now, let's show that A and B have the same rank. Remember that rank is the dimension of the column space of a matrix. To prove that similar matrices have the same rank, we need to show that the column spaces of A and B have the same dimension. Consider the equation B = P^(-1)AP. Now, let's apply a transformation to a vector x in the column space of A: \(Ax = y\) \(P^{-1}APx = P^{-1}y\) We can see that applying the matrix P^(-1)AP to a vector x in the column space of A results in a vector y in the column space of A transformed by P^(-1). This implies that the column space of B is related to the column space of A via the invertible matrix P^(-1), which is an injective linear transformation. Since P^(-1) is invertible, it maps basis vectors in the column space of A to linearly independent vectors in the column space of B, retaining the same dimension for both column spaces. Therefore, similar matrices have the same rank.

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Most popular questions from this chapter

Let \(G\) be the linear operator on \(\mathbf{R}^{3}\) defined by \(G(x, y, z)=(2 y+z, x-4 y, 3 x)\) (a) Find the matrix representation of \(G\) relative to the basis \\[ S=\left\\{w_{1}, w_{2}, w_{3}\right\\}=\\{(1,1,1), \quad(1,1,0), \quad(1,0,0)\\} \\] (b) Verify that \([G][v]=[G(v)]\) for any vector \(v\) in \(\mathbf{R}^{3}\). First find the coordinates of an arbitrary vector \((a, b, c) \in \mathbf{R}^{3}\) with respect to the basis \(S .\) Write \((a, b, c)\) as a linear combination of \(w_{1}, w_{2}, w_{3}\) using unknown scalars \(x, y,\) and \(z:\) \\[ (a, b, c)=x(1,1,1)+y(1,1,0)+z(1,0,0)=(x+y+z, x+y, x) \\] Set corresponding components equal to each other to obtain the system of equations \\[ x+y+z=a, \quad x+y=b, \quad x=c \\] Solve the system for \(x, y, z\) in terms of \(a, b, c\) to find \(x=c, y=b-c, z=a-b .\) Thus, \\[ (a, b, c)=c w_{1}+(b-c) w_{2}+(a-b) w_{3}, \quad \text { or equivalently, } \quad[(a, b, c)]=[c, b-c, a-b]^{T} \\] (a) Because \(G(x, y, z)=(2 y+z, x-4 y, 3 x)\) \\[ \begin{array}{l} G\left(w_{1}\right)=G(1,1,1)=(3,-3,3)=3 w_{1}-6 x_{2}+6 x_{3} \\ G\left(w_{2}\right)=G(1,1,0)=(2,-3,3)=3 w_{1}-6 w_{2}+5 w_{3} \\ G\left(w_{3}\right)=G(1,0,0)=(0,1,3)=3 w_{1}-2 w_{2}-w_{3} \end{array} \\] Write the coordinates \(G\left(w_{1}\right), G\left(w_{2}\right), G\left(w_{3}\right)\) as columns to get \\[ [G]=\left[\begin{array}{rrr} 3 & 3 & 3 \\ -6 & -6 & -2 \\ 6 & 5 & -1 \end{array}\right] \\] (b) Write \(G(v)\) as a linear combination of \(w_{1}, w_{2}, w_{3},\) where \(v=(a, b, c)\) is an arbitrary vector in \(\mathbf{R}^{3}\) \\[ G(v)=G(a, b, c)=(2 b+c, a-4 b, 3 a)=3 a w_{1}+(-2 a-4 b) w_{2}+(-a+6 b+c) w_{3} \\] or equivalently, \\[ [G(v)]=[3 a,-2 a-4 b,-a+6 b+c]^{T} \\] Accordingly, \\[ [G][v]=\left[\begin{array}{rrr} 3 & 3 & 3 \\ -6 & -6 & -2 \\ 6 & 5 & -1 \end{array}\right]\left[\begin{array}{c} c \\ b-c \\ a-b \end{array}\right]=\left[\begin{array}{c} 3 a \\ -2 a-4 b \\ -a+6 b+c \end{array}\right]=[G(v)] \\]

Consider the following bases of \(\mathbf{R}^{2}\) : \\[ E=\left\\{e_{1}, e_{2}\right\\}=\\{(1,0),(0,1)\\} \quad \text { and } \quad S=\left\\{u_{1}, u_{2}\right\\}=\\{(1,3),(1,4)\\} \\] (a) Find the change-of-basis matrix \(P\) from the usual basis \(E\) to \(S\). (b) Find the change-of-basis matrix \(Q\) from \(S\) back to \(E\) (c) Find the coordinate vector \([v]\) of \(v=(5,-3)\) relative to \(S\) (a) Because \(E\) is the usual basis, simply write the basis vectors in \(S\) as columns: \(P=\left[\begin{array}{ll}1 & 1 \\ 3 & 4\end{array}\right]\) (b) Method 1. Use the definition of the change-of-basis matrix. That is, express each vector in \(E\) as a linear combination of the vectors in \(S\). We do this by first finding the coordinates of an arbitrary vector \(v=(a, b)\) relative to \(S\). We have \\[ (a, b)=x(1,3)+y(1,4)=(x+y, 3 x+4 y) \quad \text { or } \quad \begin{array}{r} x+y=a \\ 3 x+4 y=b \end{array} \\] Solve for \(x\) and \(y\) to obtain \(x=4 a-b, y=-3 a+b .\) Thus, \\[ v=(4 a-b) u_{1}+(-3 a+b) u_{2} \quad \text { and } \quad[v]_{S}=[(a, b)]_{S}=[4 a-b,-3 a+b]^{T} \\] Using the above formula for \([v]_{S}\) and writing the coordinates of the \(e_{i}\) as columns yields \\[ \begin{array}{l} e_{1}=(1,0)=4 u_{1}-3 u_{2} \\ e_{2}=(0,1)=-u_{1}+u_{2} \end{array} \quad \text { and } \quad Q=\left[\begin{array}{rr} 4 & -1 \\ -3 & 1 \end{array}\right] \\] Method 2. Because \(Q=P^{-1}\), find \(P^{-1}\), say by using the formula for the inverse of a \(2 \times 2\) matrix. Thus, \\[ P^{-1}=\left[\begin{array}{rr} 4 & -1 \\ -3 & 1 \end{array}\right] \\] (c) Method 1. Write \(v\) as a linear combination of the vectors in \(S\), say by using the above formula for \(v=(a, b) .\) We have \(v=(5,-3)=23 u_{1}-18 u_{2},\) and so \([v]_{S}=[23,-18]^{T}\) Method 2. Use, from Theorem 6.6, the fact that \([v]_{S}=P^{-1}[v]_{E}\) and the fact that \([v]_{E}=[5,-3]^{T}\) \\[ [v]_{S}=P^{-1}[v]_{E}=\left[\begin{array}{rr} 4 & -1 \\ -3 & 1 \end{array}\right]\left[\begin{array}{r} 5 \\ -3 \end{array}\right]=\left[\begin{array}{r} 23 \\ -18 \end{array}\right] \\]

Suppose \(S=\left\\{u_{1}, u_{2}\right\\}\) is a basis of \(V,\) and \(T: V \rightarrow V\) is defined by \(T\left(u_{1}\right)=3 u_{1}-2 u_{2}\) and \(T\left(u_{2}\right)=u_{1}+4 u_{2}\) Suppose \(S^{\prime}=\left\\{w_{1}, w_{2}\right\\}\) is a basis of \(V\) for which \(w_{1}=u_{1}+u_{2}\) and \(w_{2}=2 u_{1}+3 u_{2}\).

Consider the following linear operator \(G\) on \(\mathbf{R}^{2}\) and basis \(S:\) \\[ G(x, y)=(2 x-7 y, 4 x+3 y) \quad \text { and } \quad S=\left\\{u_{1}, u_{2}\right\\}=\\{(1,3),(2,5)\\} \\] (a) Find the matrix representation \([G]_{S}\) of \(G\) relative to \(S\). (b) Verify \([G]_{S}[v]_{S}=[G(v)]_{S}\) for the vector \(v=(4,-3)\) in \(\mathbf{R}^{2}\) First find the coordinates of an arbitrary vector \(v=(a, b)\) in \(\mathbf{R}^{2}\) relative to the basis \(S .\) We have \\[ \left[\begin{array}{l} a \\ b \end{array}\right]=x\left[\begin{array}{l} 1 \\ 3 \end{array}\right]+y\left[\begin{array}{l} 2 \\ 5 \end{array}\right], \quad \text { and so } \quad \begin{array}{r} x+2 y=a \\ 3 x+5 y=b \end{array} \\] Solve for \(x\) and \(y\) in terms of \(a\) and \(b\) to get \(x=-5 a+2 b, y=3 a-b .\) Thus, \\[ (a, b)=(-5 a+2 b) u_{1}+(3 a-b) u_{2} \\] \\[ \text { and so } \quad[v]=[-5 a+2 b, \quad 3 a-b]^{T} \\] (a) Using the formula for \((a, b)\) and \(G(x, y)=(2 x-7 y, 4 x+3 y)\), we have \\[ \begin{array}{l} G\left(u_{1}\right)=G(1,3)=(-19,13)=121 u_{1}-70 u_{2} \\ G\left(u_{2}\right)=G(2,5)=(-31,23)=201 u_{1}-116 u_{2} \end{array} \quad \text { and so } \quad[G]_{S}=\left[\begin{array}{rr} 121 & 201 \\ -70 & -116 \end{array}\right] \\] (We emphasize that the coefficients of \(u_{1}\) and \(u_{2}\) are written as columns, not rows, in the matrix representation.) (b) Use the formula \((a, b)=(-5 a+2 b) u_{1}+(3 a-b) u_{2}\) to get Then \\[ \begin{array}{c} v=(4,-3)=-26 u_{1}+15 u_{2} \\ G(v)=G(4,-3)=(20,7)=-131 u_{1}+80 u_{2} \\ {[v]_{S}=[-26,15]^{T} \quad \text { and } \quad[G(v)]_{S}=[-131,80]^{T}} \end{array} \\] Accordingly, \\[ [G]_{S}[v]_{S}=\left[\begin{array}{rr} 121 & 201 \\ -70 & -116 \end{array}\right]\left[\begin{array}{r} -26 \\ 15 \end{array}\right]=\left[\begin{array}{r} -131 \\ 80 \end{array}\right]=[G(v)]_{S} \\] (This is expected from Theorem \(6.1 .)\)

Suppose \(F: V \rightarrow V\) is linear. A subspace \(W\) of \(V\) is said to be invariant under \(F\) if \(F(W) \subseteq W\). Suppose \(W\) is invariant under \(F\) and \(\operatorname{dim} W=r .\) Show that \(F\) has a block triangular matrix representation \(M=\left[\begin{array}{ll}A & B \\ 0 & C\end{array}\right]\).
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