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Chapter 10: Problem 69

Prove that any two projections of the same rank are similar,

Short Answer

Expert verified
In conclusion, any two projection matrices P and Q of the same rank are similar. Since both projections have the same rank, they project their vector spaces onto subspaces W(P) and W(Q) of the same dimension. An invertible matrix S exists that maps W(P) onto W(Q), and by showing that \(Q = SPS^{-1}\), we proved that P and Q are similar matrices.

Step by step solution

01

Understand Projection Matrices

A projection matrix P is a matrix that maps a vector space V onto its subspace W such that \(P^2 = P\). In other words, applying the projection matrix P twice on any vector has the same effect as applying it once. Additionally, the rank of the projection matrix P is the dimension of its target subspace W.
02

Define Similar Matrices

Two matrices A and B are similar if there exists an invertible matrix S such that \(B = S^{-1}AS\).
03

Projections with the Same Rank

If any two projection matrices P and Q have the same rank, it means they both project their vector spaces onto subspaces of the same dimension. Let's denote these subspaces as W(P) and W(Q), respectively.
04

Find the Invertible Matrix

Since W(P) and W(Q) are subspaces of the same dimension, there exists a linear transformation (and therefore an invertible matrix S) that maps W(P) onto W(Q). The inverse of this transformation is also a linear transformation, represented by the matrix \(S^{-1}\).
05

Show that P and Q are Similar

Let's now show that P and Q are similar via the invertible matrix S defined in Step 4. Since matrix S maps W(P) onto W(Q), we have: \(S^{-1}QS(Px) = S^{-1}Q(Sx) = S^{-1}(S(Px)) = Px\) for every vector in W(P). Thus, \(S^{-1}QSP = P\). Rearranging, we get the desired relation: \(Q = SPS^{-1}\). Hence, P and Q are similar matrices. In conclusion, we have proved that any two projection matrices of the same rank are similar by finding an invertible matrix S that relates them, as required.

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Most popular questions from this chapter

Let \(W\) be a subspace of \(V\). Suppose the set of cosets \(\left\\{v_{1}+W, \quad v_{2}+W, \ldots, v_{n}+W\right\\}\) in \(V / W\) is linearly independent. Show that the set of vectors \(\left\\{v_{1}, v_{2}, \ldots, v_{n}\right\\}\) in \(V\) is also linearly independent.

Let \(W\) be a substance of \(V\). Suppose the set of vectors \(\left\\{u_{1}, u_{2}, \ldots, u_{n}\right\\}\) in \(V\) is linearly independent, and that \(L\left(u_{i}\right) \cap W=\\{0\\} .\) Show that the set of cosets \(\left\\{u_{1}+W, \ldots, u_{n}+W\right\\}\) in \(V / W\) is also linearly independent.

Suppose \(V=W_{1} \oplus \cdots \oplus W_{r} .\) The projection of \(V\) into its subspace \(W_{k}\) is the mapping \(E: V \rightarrow V\) defined by \(E(v)=w_{k},\) where \(v=w_{1}+\cdots+w_{r}, w_{i} \in W_{i} .\) Show that (a) \(E\) is linear, (b) \(E^{2}=E\)

Show that \(V=W_{1} \oplus \cdots \oplus W_{r}\) if and only if (i) \(V=\operatorname{span}\left(W_{i}\right)\) and (ii) for \(k=1,2, \ldots, r\) \(W_{k} \cap \operatorname{span}\left(W_{1}, \ldots, W_{k-1}, W_{k+1}, \ldots, W_{r}\right)=\\{0\\}\).

Suppose \(\left\\{W_{i}\right\\}\) is a collection of \(T\) -invariant subspaces of a vector space \(V\). Show that the intersection \(W=\bigcap_{i} W_{i}\) is also \(T\) -invariant. Suppose \(v \in W ;\) then \(v \in W_{i}\) for every \(i .\) Because \(W_{i}\) is \(T\) -invariant, \(T(v) \in W_{i}\) for every \(i .\) Thus, \(T(v) \in W\) and so \(W\) is \(T\) -invariant.
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