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Chapter 6: Problem 16

The vectors \(u_{1}=(1,1,0), u_{2}=(0,1,1), u_{3}=(1,2,2)\) form a basis \(S\) of \(\mathbf{R}^{3}\). Find the coordinates of an arbitrary vector \(v=(a, b, c)\) relative to the basis \(S\).

Short Answer

Expert verified
The coordinates of the vector \(v = (a, b, c)\) relative to the basis \(S\) are \((b-c, -a+c, c-b+a)\).

Step by step solution

01

Write the linear combination of the basis vectors

We need to express the vector \(v = (a,b,c)\) as a linear combination of the basis vectors \(u_1 = (1,1,0), u_2 = (0,1,1),\) and \(u_3 = (1,2,2)\): $$ v = x_1 u_1 + x_2 u_2 + x_3 u_3. $$ Our goal is to find the coefficients \(x_1, x_2, x_3\).
02

Set up the system of linear equations

We can equate the components of the vector equation from Step 1: $$ a = x_1 + 0\cdot x_2 + x_3, $$ $$ b = x_1 + x_2 + 2x_3, $$ $$ c = 0\cdot x_1 + x_2 + 2x_3. $$ This gives us the following system of linear equations: $$ \begin{cases} x_1 + x_3 = a \\ x_1 + x_2 + 2x_3 = b \\ x_2 + 2x_3 = c \end{cases} $$
03

Solve the system of linear equations

We will use the substitution method to solve the system of linear equations: From the first equation, we can write \(x_1 = a - x_3\). Replace \(x_1\) in the second equation with the expression \(a - x_3\): $$ (a - x_3) + x_2 + 2x_3 = b, $$ which simplifies to \(x_2 + x_3 = b - a\). Now, replace \(x_2\) in the third equation with the expression \(b-a-x_3\): $$ (b - a - x_3) + 2x_3 = c, $$ and simplify to obtain \(x_3 = c - b + a\). Finally, we can find \(x_1\) and \(x_2\) using our previous expressions for \(x_1\) and \(x_2\): $$ x_1 = a - (c - b + a) = b - c, $$ $$ x_2 = b - a - (c - b + a) = -a + c. $$
04

Write the coordinates of the vector v

Now that we have found the coefficients \(x_1, x_2, x_3\), we can express the coordinates of the vector \(v = (a,b,c)\) relative to the basis \(S\) as \((x_1, x_2, x_3) = (b - c, -a + c, c - b + a)\).

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Most popular questions from this chapter

Let \(V\) be the vector space of \(2 \times 2\) matrices. Consider the following matrix \(M\) and usual basis \(E\) of \(V\) \(M=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\) and \(\quad E=\left\\{\left[\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right],\left[\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right], \quad\left[\begin{array}{ll}0 & 0 \\ 1 & 0\end{array}\right], \quad\left[\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right]\right\\}\) Find the matrix representing each of the following linear operators \(T\) on \(V\) relative to \(E\) (a) \(T(A)=M A\) (b) \(T(A)=A M\) (c) \(T(A)=M A-A M\)

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 \(M=\left[\begin{array}{ll}A & B \\ 0 & C\end{array}\right]\).

Let \(1_{V}\) and \(0_{V}\) denote the identity and zero operators, respectively, on a vector space \(V\). Show that, for any basis \(S\) of \(V\) (a) \(\left[\mathbf{1}_{V}\right]_{S}=I,\) the identity matrix. (b) \(\left[\mathbf{0}_{V}\right]_{S}=0,\) the zero matrix.

Let \(A: \mathbf{R}^{2} \rightarrow \mathbf{R}^{2}\) be defined by the matrix \(A=\left[\begin{array}{rr}5 & -1 \\ 2 & 4\end{array}\right]\).

Let \(S\) and \(S^{\prime}\) be bases of \(V\), and let \(1_{V}\) be the identity mapping on \(V\). Show that the matrix \(A\) representing \(\mathbf{1}_{V}\) relative to the bases \(S\) and \(S^{\prime}\) is the inverse of the change-of- basis matrix \(P\) from \(S\) to \(S^{\prime} ;\) that is, \(A=P^{-1}\).
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