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Chapter 10: Problem 68

Suppose \(E: V \rightarrow V\) is a projection (i.e., \(E^{2}=E\) ). Prove that \(E\) has a matrix representation of the form \(\left[\begin{array}{cc}I_{r} & 0 \\\ 0 & 0\end{array}\right],\) where \(r\) is the rank of \(E\) and \(I_{r}\) is the \(r\) -square identity matrix.

Short Answer

Expert verified
The matrix representation of the projection operator \(E\) with respect to the basis \(B\) obtained from the direct sum decomposition of the vector space \(V\) is given by \(\left[\begin{array}{cc} I_r & 0 \\ 0 & 0 \end{array}\right]\), where \(r\) is the rank of \(E\) and \(I_r\) is the \(r \times r\) identity matrix.

Step by step solution

01

Decompose V into a direct sum of subspaces

We need to find subspaces \(W_1\) and \(W_2\) of \(V\) such that \(V = W_1 \oplus W_2\). Let \(W_1 = \operatorname{Im}(E)\) (the image of \(E\)) and \(W_2 = \ker(E)\) (the kernel of \(E\)). We will prove that any vector \(v \in V\) can be written uniquely as \(v = w_1 + w_2\), where \(w_1 \in W_1\) and \(w_2 \in W_2\). Consider \(v = Ev + (v - Ev)\). Clearly, \(Ev \in W_1\) since it is in the image of \(E\), and \((v - Ev) \in W_2\) because \(E(v - Ev) = E(v) - E^2(v) = E(v) - E(v) = 0\). Thus, we can write \(v\) as the sum of a vector from \(W_1\) and a vector from \(W_2\). It remains to show that this decomposition is unique. Suppose \(v = w_1' + w_2'\) for some \(w_1' \in W_1\) and \(w_2' \in W_2\). Then \(w_1' = Ew_1'\), and since \(ew_2' \in W_2\) and \(W_1\) and \(W_2\) are both subspaces, we can apply the projection operator \(E\) to both sides of the equation to get $$E(v) = E(w_1' + w_2') = Ew_1' + Ew_2'.$$ From this, we can obtain \(w_1 = w_1'\) and \(w_2 = w_2'\). Thus, the decomposition is unique, and therefore \(V = W_1 \oplus W_2\).
02

Choose basis for V

Let \(B_1 = \{v_1, \dots, v_r\}\) be a basis for \(W_1 = \operatorname{Im}(E)\) and \(B_2 = \{v_{r+1},\dots, v_n \}\) be a basis for \(W_2 = \ker(E)\). Then, the union of these bases \(B = B_1 \cup B_2 = \{v_1, \dots, v_n\}\) forms a basis for \(V\), since every vector \(v\) in \(V\) can be expressed uniquely as the linear combination of vectors in \(B_1\) and \(B_2\).
03

Matrix representation of E

Now let's write down the matrix representation of \(E\) with respect to the basis \(B\). Recall that the columns of this matrix are just the images of the basis vectors \(E(v_j)\) expressed as linear combinations of \(B\). We have $$E(v_j) = \left\{\begin{array}{ll} v_j & \;\; \text{for } 1 \leq j \leq r \\ 0 & \;\; \text{for } r+1 \leq j \leq n \end{array}\right..$$ Therefore, using the basis \(B\), the matrix representation of \(E\) is given by $$ \left[\begin{array}{ccc|ccc} 1 & & & & & \\ & \ddots & & & 0 & \\ & & 1 & & & \\ \hline & & & 0 & & \\ & 0 & & & \ddots & \\ & & & & & 0 \end{array}\right], $$ which is in the desired form \(\left[\begin{array}{cc} I_r & 0 \\ 0 & 0 \end{array}\right]\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Algebra
Linear algebra is a fundamental branch of mathematics that studies vectors, vector spaces, linear mappings, and systems of linear equations. It is invaluable in various fields such as physics, economics, engineering, computer science, and more. In essence, linear algebra is about understanding spaces constituted by vectors and the transformations that operate within these spaces.

When it comes to projection matrices like the one in the exercise, we are dealing with a type of linear transformation that maps a vector space onto a subspace. The properties of these transformations, particularly their representation through matrices, are key to solving linear algebra problems. Thinking in terms of matrices and their operations simplifies complex problems, and the representation of a projection matrix with a form such as \(\left[\begin{array}{cc}I_{r} & 0\0 & 0\end{array}\right]\) is a classic result that showcases the power of matrix algebra in expressing linear transformations.
Direct Sum of Subspaces
In linear algebra, the concept of the direct sum of subspaces provides a structured way to piece together vector spaces. When you decompose a vector space \( V \) into a direct sum of subspaces \( V = W_1 \oplus W_2 \), as seen in the exercise, it is like separating the space into two distinct, non-overlapping parts. Each vector in \( V \) can be uniquely written as the sum of vectors from each subspace. This idea is crucial in many applications, such as simplifying linear maps and studying vibrations in mechanical systems.

Understanding direct sums is also essential for solving systems of linear equations. It determines how to combine solutions from different parts without interference, like playing two different notes on a piano to create a harmony that is richer than the individual sounds. For a projection operator \( E \) in the exercise, it means breaking down the space so that \( E \) acts like the identity on one part and like a zero on another鈥攖he independence of the subspaces ensures that the projection operator 'projects' vectors correctly onto the desired subspace.
Basis of Vector Space
The concept of a basis of a vector space is like the DNA of that space. In linear algebra, a basis is essentially a set of vectors that are both linearly independent and span the vector space鈥攚hich means you can combine them in various ways to produce any vector in the space. In the context of the exercise, establishing a basis \( B \) for the vector space \( V \) allows us to express any linear transformation, including projections, in a clear, manageable form through matrix representation.

Choosing a basis \( B \) as the union of bases for \( W_1 \) and \( W_2 \) provides a foundation on which the entire vector space can be understood and manipulated. Each vector in \( V \) has a unique representation in terms of the basis vectors, making calculations, such as those needed for finding the matrix representation of \( E \), much more straightforward. The importance of using a basis for problem-solving in linear algebra cannot be overstated鈥攊t is like having a map in unknown terrain, guiding you through your calculations to the desired destination: a solution.

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Most popular questions from this chapter

Let \(\hat{T}\) denote the restriction of an operator \(T\) to an invariant subspace \(W\). Prove (a) For any polynomial \(f(t), f(\hat{T})(w)=f(T)(w)\) (b) The minimal polynomial of \(\hat{T}\) divides the minimal polynomial of \(T\) (a) If \(f(t)=0\) or if \(f(t)\) is a constant (i.e., of degree 1 ), then the result clearly holds. Assume deg \(f=n>1\) and that the result holds for polynomials of degree less than \(n\). Suppose that Then $$\begin{aligned} & f(t)=a_{n} t^{n}+a_{n-1} t^{n-1}+\cdots+a_{1} t+a_{0} \\ f(\hat{T})(w) &=\left(a_{n} \hat{T}^{n}+a_{n-1} \hat{T}^{n-1}+\cdots+a_{0} I\right)(w) \\ &=\left(a_{n} \hat{T}^{n-1}\right)(\hat{T}(w))+\left(a_{n-1} \hat{T}^{n-1}+\cdots+a_{0} I\right)(w) \\ &=\left(a_{n} T^{n-1}\right)(T(w))+\left(a_{n-1} T^{n-1}+\cdots+a_{0} I\right)(w)=f(T)(w) \end{aligned}$$ (b) Let \(m(t)\) denote the minimal polynomial of \(T .\) Then by (a), \(m(\hat{T})(w)=m(T)(w)=\mathbf{0}(w)=0\) for every \(w \in W ;\) that is, \(\hat{T}\) is a zero of the polynomial \(m(t) .\) Hence, the minimal polynomial of \(T\) divides \(m(t)\)

Suppose \(T_{1}\) and \(T_{2}\) are nilpotent operators that commute (i.e., \(T_{1} T_{2}=T_{2} T_{1}\) ). Show that \(T_{1}+T_{2}\) and \(T_{1} T_{2}\) are also nilpotent.

Find all possible Jordan canonical forms for those matrices whose characteristic polynomial \(\Delta(t)\) and minimal polynomial \(m(t)\) are as follows: (a) \(\Delta(t)=(t-2)^{4}(t-3)^{2}, m(t)=(t-2)^{2}(t-3)^{2}\) (b) \(\Delta(t)=(t-7)^{5}, m(t)=(t-7)^{2}\) (c) \(\Delta(t)=(t-2)^{7}, m(t)=(t-2)^{3}\)

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

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 \(k=1,2, \ldots, r\) \(W_{k} \cap \operatorname{span}\left(W_{1}, \ldots, W_{k-1}, W_{k+1}, \ldots, W_{r}\right)=\\{0\\}\).
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