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Math
Posted 8 months ago

Q4[5\mathbf{Q 4}\left[5\right. marks] Let V=(V,+,,R)V=(V,+, \cdot, \mathbb{R}) be a vector space and let B={v1,v2,v3}VB=\left\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}\right\} \subseteq V be a basis.
a) Define what it means that the triple (a1,a2,a3)R3\left(a_{1}, a_{2}, a_{3}\right) \in \mathbb{R}^{3} is the coordinate vector of xV\mathbf{x} \in V with respect to the basis BB.
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School of Mathematics and Statistics
University of New South Wales
b) Find the coordinate vector of v1\mathbf{v}_{1} with respect to the basis BB.
c) Define what it means that a matrix AM3,3(R)A \in M_{3,3}(\mathbb{R}) is the matrix of the linear map T:VVT: V \rightarrow V with respect to the basis BB.
d) Let
A=(010100000).
A=\left(\begin{array}{lll}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 0
\end{array}\right) .

be the matrix of the map TT with respect to the basis BB. Find T(v1)T\left(\mathbf{v}_{1}\right) in terms of BB.
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Answer from Sia
Posted 8 months ago
Solution by Steps
step 1
The coordinate vector of xV\mathbf{x} \in V with respect to the basis B={v1,v2,v3}B = \{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\} is a unique ordered triple (a1,a2,a3)R3(a_1, a_2, a_3) \in \mathbb{R}^3 such that
step 2
x=a1v1+a2v2+a3v3\mathbf{x} = a_1 \mathbf{v}_1 + a_2 \mathbf{v}_2 + a_3 \mathbf{v}_3


 Part (b)
step 1
The coordinate vector of v1\mathbf{v}_1 with respect to the basis BB is found by expressing v1\mathbf{v}_1 as a linear combination of the basis vectors
step 2
Since v1\mathbf{v}_1 is itself a basis vector, its coordinate vector is (1,0,0)(1, 0, 0)


 Part (c)
step 1
A matrix AM3,3(R)A \in M_{3,3}(\mathbb{R}) is the matrix of the linear map T:VVT: V \rightarrow V with respect to the basis BB if
step 2
AA is the matrix whose columns are the coordinate vectors of T(v1),T(v2),T(v3)T(\mathbf{v}_1), T(\mathbf{v}_2), T(\mathbf{v}_3) with respect to BB


 Part (d)
step 1
Given the matrix A=(010100000)A = \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}, we need to find T(v1)T(\mathbf{v}_1) in terms of BB
step 2
The first column of AA represents the coordinate vector of T(v1)T(\mathbf{v}_1) with respect to BB
step 3
Therefore, T(v1)=0v1+1v2+0v3=v2T(\mathbf{v}_1) = 0 \mathbf{v}_1 + 1 \mathbf{v}_2 + 0 \mathbf{v}_3 = \mathbf{v}_2
Answer
T(v1)=v2T(\mathbf{v}_1) = \mathbf{v}_2
Key Concept
Coordinate Vector
Explanation
The coordinate vector of a vector x\mathbf{x} with respect to a basis BB is the unique ordered triple that expresses x\mathbf{x} as a linear combination of the basis vectors.

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