Problem 38 - Differential Equations and Linear Algebra

Question

In this problem, we establish that similar matrices describe the same linear transformation relative to different bases. Assume that \(\left\\{\mathbf{e}_{1}, \mathbf{e}_{2}, \ldots, \mathbf{e}_{n}\right\\}\) and \(\left\\{\mathbf{f}_{1}, \mathbf{f}_{2}, \ldots, \mathbf{f}_{n}\right\\}\) are bases for a vector space \(V\) and let \(T: V \rightarrow V\) be a linear transformation. Define the \(n \times n\) matrices \(A=\left[a_{i k}\right]\) and \(B=\left[b_{i k}\right]\) by $$\begin{aligned} &T\left(\mathbf{e}_{k}\right)=\sum_{i=1}^{n} a_{i l} \mathbf{e}_{i}, \quad k=1,2, \ldots, n\\\ &T\left(\mathbf{f}_{k}\right)=\sum_{i=1}^{n} b_{i k} \mathbf{f}_{i}, \quad k=1,2, \ldots, n \end{aligned}$$ If we express each of the basis vectors \(\mathbf{f}_{1}, \mathbf{f}_{2}, \ldots, \mathbf{f}_{n}\) in terms of the basis vectors \(\mathbf{e}_{1}, \mathbf{e}_{2}, \ldots, \mathbf{e}_{n},\) we have that $$\mathbf{f}_{i}=\sum_{j=1}^{n} s_{j i} \mathbf{e}_{j}, \quad i=1,2, \ldots, n$$ for appropriate scalars \(s_{j i} .\) Thus, the matrix \(S=\left[s_{j i}\right]\) describes the relationship between the two bases. (a) Prove that \(S\) is nonsingular. [Hint: Use the fact that \(\left.\mathbf{f}_{1}, \mathbf{f}_{2}, \ldots, \mathbf{f}_{n} \text { are linearly independent. }\right]\) (b) Use \((7.3 .14)\) and \((7.3 .15)\) to show that, for \(k=1,2, \ldots, n,\) we haveor equivalently, $$T\left(\mathbf{f}_{k}\right)=\sum_{i=1}^{n}\left(\sum_{j=1}^{n} s_{i j} b_{j k}\right) \mathbf{e}_{i}$$ (c) Use \((7.3 .13)\) and \((7.3 .15)\) to show that, for \(k=\) \(1,2, \ldots n,\) we have $$T\left(\mathbf{f}_{k}\right)=\sum_{i=1}^{n}\left(\sum_{j=1}^{n} a_{i j} s_{j k}\right) \mathbf{e}_{i}$$ (d) Use \((7.3 .16)\) and \((7.3 .17)\) together with the fact that \(\left\\{\mathbf{e}_{1}, \mathbf{e}_{2}, \ldots, \mathbf{e}_{n}\right\\}\) is linearly independent to show that $$\sum_{j=1}^{n} s_{i j} b_{j k}=\sum_{j=1}^{n} a_{i j} s_{j k}, \quad 1 \leq i, k \leq n$$ and hence that $$S B=A S$$ Finally, conclude that \(A\) and \(B\) are related by $$B=S^{-1} A S$$.

Step-by-Step Solution

Answer

In summary, we have shown that the two matrices A and B describing the same linear transformation with respect to different bases are related by the expression \(B = S^{-1}AS\). To reach this result, we first proved that the matrix S, representing the relationship between the two bases, is nonsingular. Following that, we found expressions for the linear transformation T(𝑓ₖ) in terms of A, B, and S. By setting these expressions equal to each other and utilizing the linear independence of the basis, we derived the desired relationship between A and B.

Step 1: Showing that S is nonsingular

S describes the relationship between two bases, which are linearly independent sets of vectors. To show that S is nonsingular, we need to prove that the transformation S is invertible. Also, S is invertible if |S| ≠ 0. Therefore, we need to show that |S| ≠ 0. Let's assume to the contrary that |S| = 0. Then, there exists a nonzero vector 𝑥 with components x₁, x₂, ... , xₙ such that Sx = 0. This would mean that: \(\sum_{j=1}^{n} s_{j i} x_{j} = \mathbf{0} \text{ for } i = 1,2, \ldots, n\) Now consider the vector \(w = \sum_{i=1}^{n} x_{i} \mathbf{f}_{i}\). Since \(\mathbf{f}_{i} = \sum_{j=1}^{n} s_{j i} \mathbf{e}_{j}\), we can rewrite each \(\mathbf{f}_{i}\) in terms of \(\mathbf{e}_{i}\). \(w = \sum_{i=1}^{n} x_{i}\sum_{j=1}^{n} s_{j i} \mathbf{e}_{j} = \sum_{j=1}^{n}\left(\sum_{i=1}^{n} x_{i} s_{j i}\right) \mathbf{e}_{j}\) However, according to our assumption, Sx = 0, meaning each sum inside the parenthesis is zero. This implies w = 0. But since the basis \(\{\mathbf{f}_{1}, \mathbf{f}_{2}, \ldots, \mathbf{f}_{n}\}\) is linearly independent and x is nonzero, this contradiction shows that our assumption is wrong, and |S| ≠ 0. Therefore, S is nonsingular. #Part (b)#

Step 2: Expression for T(𝑓𝑘) in terms of A and B

We are given the expression for T(𝑓ₖ) in terms of B: \(T\left(\mathbf{f}_{k}\right)=\sum_{i=1}^{n} b_{i k} \mathbf{f}_{i}\). We want to find the same expression but in terms of the basis \(\mathbf{e}_{i}\) instead of \(\mathbf{f}_{i}\). We know that: \(\mathbf{f}_{i}=\sum_{j=1}^{n} s_{j i} \mathbf{e}_{j}\) Now, substituting this expression into the expression for T(𝑓ₖ) in terms of B, we get: \(T\left(\mathbf{f}_{k}\right) = \sum_{i=1}^{n} b_{i k} \left(\sum_{j=1}^{n} s_{j i} \mathbf{e}_{j}\right)\) By switching the summation order, we obtain: \(T\left(\mathbf{f}_{k}\right) = \sum_{i=1}^{n}\left(\sum_{j=1}^{n} s_{i j} b_{j k}\right) \mathbf{e}_{i}\) #Part (c)#

Step 3: Expression for T(𝑓𝑘) in terms of A and S

We are given the expression for T(𝑒ₖ) in terms of A: \(T\left(\mathbf{e}_{k}\right)=\sum_{i=1}^{n} a_{i k} \mathbf{e}_{i}\) We know that: \(\mathbf{f}_{k}=\sum_{j=1}^{n} s_{j k} \mathbf{e}_{j}\) So, applying T to both sides we have: \(T\left(\mathbf{f}_{k}\right) = T\left(\sum_{j=1}^{n} s_{j k} \mathbf{e}_{j}\right)\) Since T is linear, we have: \(T\left(\mathbf{f}_{k}\right)=\sum_{j=1}^{n} s_{j k} T\left(\mathbf{e}_{j}\right)\) Now substituting the given expression for T(𝑒ₖ) in terms of A: \(T\left(\mathbf{f}_{k}\right)=\sum_{j=1}^{n} s_{j k}\left(\sum_{i=1}^{n} a_{i j} \mathbf{e}_{i}\right)\) By switching the summation order, we obtain: \(T\left(\mathbf{f}_{k}\right)=\sum_{i=1}^{n}\left(\sum_{j=1}^{n} a_{i j} s_{j k}\right) \mathbf{e}_{i}\) #Part (d)#

Step 4: Showing the relationship between A and B

From parts (b) and (c), we have these two expressions for T(𝑓ₖ): \(T\left(\mathbf{f}_{k}\right)=\sum_{i=1}^{n}\left(\sum_{j=1}^{n} s_{i j} b_{j k}\right) \mathbf{e}_{i}\) \(T\left(\mathbf{f}_{k}\right)=\sum_{i=1}^{n}\left(\sum_{j=1}^{n} a_{i j} s_{j k}\right) \mathbf{e}_{i}\) Since these are two expressions for the same output, we can equate the sums: \(\sum_{j=1}^{n} s_{i j} b_{j k} = \sum_{j=1}^{n} a_{i j} s_{j k}, \quad 1 \leq i, k \leq n\) Since the basis \(\{\mathbf{e}_{1}, \mathbf{e}_{2}, \ldots, \mathbf{e}_{n}\}\) is linearly independent, we can form a system of linear equations that has the solution: $$SB = AS$$ Thus, we obtain the relationship between A and B as follows: $$B = S^{-1}AS$$ since S is nonsingular as demonstrated in step a.

Key Concepts

Linear Transformation

A linear transformation is a function between two vector spaces that preserves the operations of vector addition and scalar multiplication. If we denote a linear transformation as \(T: V \rightarrow V\), it maps vectors from the vector space \(V\) to itself in such a way that if \(\mathbf{u}\) and \(\mathbf{v}\) are vectors in \(V\) and \(c\) is a scalar, then two key properties hold:

• \(T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})\)
• \(T(c\mathbf{u}) = cT(\mathbf{u})\)

These properties ensure that the structure of the vector space is maintained. In our problem, we are interested in understanding how a linear transformation \(T\) can be expressed in terms of two different bases \(\{\mathbf{e}_1, \mathbf{e}_2, \ldots, \mathbf{e}_n\}\) and \(\{\mathbf{f}_1, \mathbf{f}_2, \ldots, \mathbf{f}_n\}\). The matrices \(A\) and \(B\) are used to represent the transformation \(T\) with respect to these different bases.

Vector Space

A vector space is a collection of objects called vectors, which can be added together and multiplied by scalars, maintaining the operation's properties like closure, associativity, commutativity, etc. Formally, if \(V\) is a vector space over a field \(F\), it must satisfy the following axioms:

• Vector addition is commutative: \(\mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u}\)
• Vector addition is associative: \((\mathbf{u} + \mathbf{v}) + \mathbf{w} = \mathbf{u} + (\mathbf{v} + \mathbf{w})\)
• There is an additive identity: a zero vector \(\mathbf{0}\) such that \(\mathbf{u} + \mathbf{0} = \mathbf{u}\)
• Each vector has an additive inverse: a vector \(-\mathbf{u}\) such that \(\mathbf{u} + (-\mathbf{u}) = \mathbf{0}\)
• Scalar multiplication is distributive: \(c(\mathbf{u} + \mathbf{v}) = c\mathbf{u} + c\mathbf{v}\) and \((c+d)\mathbf{u} = c\mathbf{u} + d\mathbf{u}\)

This structure allows us to understand and manipulate vectors in a meaningful way. In the context of similar matrices, we are dealing with vector spaces that have bases, meaning that every vector in the space can be uniquely expressed as a linear combination of the basis vectors.

Change of Basis

The change of basis is the process of switching from one set of basis vectors to another within the same vector space. This is crucial because the representation of a linear transformation can look quite different when expressed in different bases. If \(\{\mathbf{e}_1, \mathbf{e}_2, \ldots, \mathbf{e}_n\}\) is one basis, and \(\{\mathbf{f}_1, \mathbf{f}_2, \ldots, \mathbf{f}_n\}\) is another, then there is a matrix \(S\) that describes how to convert coordinates from the \(\mathbf{f}\)-basis to the \(\mathbf{e}\)-basis.

• Each \(\mathbf{f}_i\) can be expressed as a combination of the \(\mathbf{e}_j\):
  \(\mathbf{f}_i = \sum_{j=1}^{n} s_{j i} \mathbf{e}_{j}\)

The matrix \(S\) with entries \(s_{ji}\) is called the change of basis matrix. In the problem, it's shown that this matrix \(S\) must be nonsingular, meaning it has an inverse \(S^{-1}\), which is necessary for the transformations between bases to be reversible.

Matrix Inversion

Matrix inversion is a technique used to find a matrix that reverses the effect of a given square matrix. For a matrix \(M\), if there exists another matrix \(N\) such that \(MN = NM = I\), where \(I\) is the identity matrix, then \(N\) is called the inverse of \(M\), denoted \(M^{-1}\).

• A matrix is invertible (or nonsingular) if and only if its determinant is non-zero.
• In the context of similar matrices, if \(S\) describes the change of basis, then \(S\) must be invertible to establish the relationship \(B = S^{-1}AS\).

The steps in the problem carefully establish \(S\) as nonsingular by demonstrating that it maps one basis to another without losing information (i.e., it doesn't collapse vectors to zero), thus \(|S| e 0\). This allows us to rewrite a linear transformation from one basis to another using similar matrices, highlighting how \(A\) and \(B\) are connected through \(S\) and its inverse.