4. Let V be a vector space, B a basis of V and T a linear...

Name: 4. Let V be a vector space, B a basis of V and T a linear operator on V. Show that m[T]B = mT.
Uploaded: 2023-08-06T19:21:54-08:00
Channel: Victor Salazar
Description: 4. Let V be a vector space, B a basis of V and T a linear operator on V. Show that m[T]B = mT.

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Let $\left\{v_{1}, v_{2}, \ldots, v_{n}\right\}$ be a basis for a vector space $V,$ and let $T: V \rightarrow V$ be a linear operator. Prove that if $$ T\left(\mathbf{v}_{1}\right)=\mathbf{v}_{1}, \quad T\left(\mathbf{v}_{2}\right)=\mathbf{v}_{2}, \ldots, \quad T\left(\mathbf{v}_{n}\right)=\mathbf{v}_{n}$$ then $T$ is the identity transformation on $V$.

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Let V be an n-dimensional vector space and let W be an m-dimensional vector space, both over the same field F, and having bases Bv = {a1, a2, ..., an} and Bw = {B1, B2, ..., Bm}. Let X = [a] Bv be the coordinate matrix for an arbitrary alpha, and let A = [aij] be any m x n matrix over the field F. Prove that the function T : V -> W defined according to the equation T(sum(xj aj)) = sum(sum(aij xj)) bi is a linear transformation.

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Find the matrix of the given linear transformation $T$ with respect to the given basis. If no basis is specified, use the standard basis: $2 \mathrm{x}=\left(1, t, t^{2}\right)$ for $P_{2}$ $$\mathfrak{A}=\left(\left[\begin{array}{ll}1 & 0 \\0 & 0\end{array}\right],\left[\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right],\left[\begin{array}{ll}0 & 0 \\1 & 0\end{array}\right],\left[\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right]\right)$$ for $\mathbb{R}^{2 \times 2},$ and $\mathfrak{A}=(1, i)$ for $\mathbb{C} .$ For the space $U^{2 \times 2}$ of upper triangular $2 \times 2$ matrices, use the basis $$\mathfrak{A}=\left(\left[\begin{array}{ll}1 & 0 \\0 & 0\end{array}\right],\left[\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right],\left[\begin{array}{ll}0 & 0 \\0 & 1\end{array}\right]\right)$$ unless another basis is given. In each case, determine whether $T$ is an isomorphism. If $T$ isn't an isomorphism, find bases of the kernel and image of $T,$ and thus deter mine the rank of $T$. $T(M)=\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right] M-M\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right]$ from $\mathbb{R}^{2 \times 2}$ to $\mathbb{R}^{2 \times 2}$ with respect to the basis $$\mathfrak{B}=\left(\left[\begin{array}{rr} 1 & 1 \\ -1 & -1 \end{array}\right],\left[\begin{array}{rr} 1 & -1 \\ 1 & -1 \end{array}\right],\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right],\left[\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right]\right)$$

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Transcript

00:01 For this exercise, we have that t v sub i equals v sub i for all i is an element of 1 to n.

00:17 And then we have for all v is an element of v.

00:29 There exists a sub 0, a sub 1, to a sub n for all r space.

00:42 Then v equals a sub 1, v sub 1 plus a sub 2, v sub 2 plus a sub n, v sub n.

00:52 Thus t v is equal to t a sub 1, v sub 1 plus a sub 2, v sub 2 plus a sub n, v sub n...

Question

4. Let V be a vector space, B a basis of V and T a linear operator on V. Show that $m_{[T]_B} = m_T$.

Question

4. Let V be a vector space, B a basis of V and T a linear operator on V. Show that $m_{[T]_B} = m_T$.

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