Find $beta _{1}, beta _{2}, beta _{3}, beta _{4} in mathbb R$
$begingroup$
In the linear space $mathbb R^{2,2}$ there are matrices $A_{1}={begin{bmatrix}1&1\1&0end{bmatrix}}$, $A_{2}={begin{bmatrix}0&1\0&1end{bmatrix}}$, $A_{3}={begin{bmatrix}1&1\0&0end{bmatrix}}$, $A_{4}={begin{bmatrix}1&0\1&1end{bmatrix}}$.
(a)Prove that $(A_{1}, A_{2}, A_{3}, A_{4})$ is a basis of the space $mathbb R^{2,2}$.
(b) $g_{1}^{*}, g_{2}^{*}, g_{3}^{*}, g_{4}^{*} in (mathbb R^{2,2})^{*}$ is a dual basis from basis $(A_{1}, A_{2}, A_{3}, A_{4})$.
Functional $f^{*} in (mathbb R^{2,2})^{*}$ is given by the formula: $f^{*}({begin{bmatrix} a_{11} & a_{12} \ a_{21} & a_{22} end{bmatrix}})=a_{11} + a_{12} + a_{21} + a_{22}$.
Find $beta _{1}, beta _{2}, beta _{3}, beta _{4} in mathbb R$ such that $f^{*}=beta _{1} g_{1}^{*}+beta _{2} g_{2}^{*}+beta _{3} g_{3}^{*}+beta _{4} g_{4}^{*}$.
I think (a) is easy and I know how to do it.
Hovewer in (b) I have my sollution but I think it is not good answer, so I need an opinion to my approach:
1) I find a dual basis from basis $(A_{1}, A_{2}, A_{3}, A_{4})$: $${begin{bmatrix}a&b\c&dend{bmatrix}}=alpha_{1} A_{1}+alpha_{2} A_{2}+alpha_{3} A_{3}+alpha_{4} A_{4}$$
2) Then I have $$alpha_{1}=g_{1}^{*}, alpha_{2}=g_{2}^{*}, alpha_{3}=g_{3}^{*}, alpha_{4}=g_{4}^{*}$$
3) $f^{*}=beta _{1} g_{1}^{*}+beta _{2} g_{2}^{*}+beta _{3} g_{3}^{*}+beta _{4} g_{4}^{*}$ so I substitute the early calculated values.
linear-algebra
$endgroup$
add a comment |
$begingroup$
In the linear space $mathbb R^{2,2}$ there are matrices $A_{1}={begin{bmatrix}1&1\1&0end{bmatrix}}$, $A_{2}={begin{bmatrix}0&1\0&1end{bmatrix}}$, $A_{3}={begin{bmatrix}1&1\0&0end{bmatrix}}$, $A_{4}={begin{bmatrix}1&0\1&1end{bmatrix}}$.
(a)Prove that $(A_{1}, A_{2}, A_{3}, A_{4})$ is a basis of the space $mathbb R^{2,2}$.
(b) $g_{1}^{*}, g_{2}^{*}, g_{3}^{*}, g_{4}^{*} in (mathbb R^{2,2})^{*}$ is a dual basis from basis $(A_{1}, A_{2}, A_{3}, A_{4})$.
Functional $f^{*} in (mathbb R^{2,2})^{*}$ is given by the formula: $f^{*}({begin{bmatrix} a_{11} & a_{12} \ a_{21} & a_{22} end{bmatrix}})=a_{11} + a_{12} + a_{21} + a_{22}$.
Find $beta _{1}, beta _{2}, beta _{3}, beta _{4} in mathbb R$ such that $f^{*}=beta _{1} g_{1}^{*}+beta _{2} g_{2}^{*}+beta _{3} g_{3}^{*}+beta _{4} g_{4}^{*}$.
I think (a) is easy and I know how to do it.
Hovewer in (b) I have my sollution but I think it is not good answer, so I need an opinion to my approach:
1) I find a dual basis from basis $(A_{1}, A_{2}, A_{3}, A_{4})$: $${begin{bmatrix}a&b\c&dend{bmatrix}}=alpha_{1} A_{1}+alpha_{2} A_{2}+alpha_{3} A_{3}+alpha_{4} A_{4}$$
2) Then I have $$alpha_{1}=g_{1}^{*}, alpha_{2}=g_{2}^{*}, alpha_{3}=g_{3}^{*}, alpha_{4}=g_{4}^{*}$$
3) $f^{*}=beta _{1} g_{1}^{*}+beta _{2} g_{2}^{*}+beta _{3} g_{3}^{*}+beta _{4} g_{4}^{*}$ so I substitute the early calculated values.
linear-algebra
$endgroup$
add a comment |
$begingroup$
In the linear space $mathbb R^{2,2}$ there are matrices $A_{1}={begin{bmatrix}1&1\1&0end{bmatrix}}$, $A_{2}={begin{bmatrix}0&1\0&1end{bmatrix}}$, $A_{3}={begin{bmatrix}1&1\0&0end{bmatrix}}$, $A_{4}={begin{bmatrix}1&0\1&1end{bmatrix}}$.
(a)Prove that $(A_{1}, A_{2}, A_{3}, A_{4})$ is a basis of the space $mathbb R^{2,2}$.
(b) $g_{1}^{*}, g_{2}^{*}, g_{3}^{*}, g_{4}^{*} in (mathbb R^{2,2})^{*}$ is a dual basis from basis $(A_{1}, A_{2}, A_{3}, A_{4})$.
Functional $f^{*} in (mathbb R^{2,2})^{*}$ is given by the formula: $f^{*}({begin{bmatrix} a_{11} & a_{12} \ a_{21} & a_{22} end{bmatrix}})=a_{11} + a_{12} + a_{21} + a_{22}$.
Find $beta _{1}, beta _{2}, beta _{3}, beta _{4} in mathbb R$ such that $f^{*}=beta _{1} g_{1}^{*}+beta _{2} g_{2}^{*}+beta _{3} g_{3}^{*}+beta _{4} g_{4}^{*}$.
I think (a) is easy and I know how to do it.
Hovewer in (b) I have my sollution but I think it is not good answer, so I need an opinion to my approach:
1) I find a dual basis from basis $(A_{1}, A_{2}, A_{3}, A_{4})$: $${begin{bmatrix}a&b\c&dend{bmatrix}}=alpha_{1} A_{1}+alpha_{2} A_{2}+alpha_{3} A_{3}+alpha_{4} A_{4}$$
2) Then I have $$alpha_{1}=g_{1}^{*}, alpha_{2}=g_{2}^{*}, alpha_{3}=g_{3}^{*}, alpha_{4}=g_{4}^{*}$$
3) $f^{*}=beta _{1} g_{1}^{*}+beta _{2} g_{2}^{*}+beta _{3} g_{3}^{*}+beta _{4} g_{4}^{*}$ so I substitute the early calculated values.
linear-algebra
$endgroup$
In the linear space $mathbb R^{2,2}$ there are matrices $A_{1}={begin{bmatrix}1&1\1&0end{bmatrix}}$, $A_{2}={begin{bmatrix}0&1\0&1end{bmatrix}}$, $A_{3}={begin{bmatrix}1&1\0&0end{bmatrix}}$, $A_{4}={begin{bmatrix}1&0\1&1end{bmatrix}}$.
(a)Prove that $(A_{1}, A_{2}, A_{3}, A_{4})$ is a basis of the space $mathbb R^{2,2}$.
(b) $g_{1}^{*}, g_{2}^{*}, g_{3}^{*}, g_{4}^{*} in (mathbb R^{2,2})^{*}$ is a dual basis from basis $(A_{1}, A_{2}, A_{3}, A_{4})$.
Functional $f^{*} in (mathbb R^{2,2})^{*}$ is given by the formula: $f^{*}({begin{bmatrix} a_{11} & a_{12} \ a_{21} & a_{22} end{bmatrix}})=a_{11} + a_{12} + a_{21} + a_{22}$.
Find $beta _{1}, beta _{2}, beta _{3}, beta _{4} in mathbb R$ such that $f^{*}=beta _{1} g_{1}^{*}+beta _{2} g_{2}^{*}+beta _{3} g_{3}^{*}+beta _{4} g_{4}^{*}$.
I think (a) is easy and I know how to do it.
Hovewer in (b) I have my sollution but I think it is not good answer, so I need an opinion to my approach:
1) I find a dual basis from basis $(A_{1}, A_{2}, A_{3}, A_{4})$: $${begin{bmatrix}a&b\c&dend{bmatrix}}=alpha_{1} A_{1}+alpha_{2} A_{2}+alpha_{3} A_{3}+alpha_{4} A_{4}$$
2) Then I have $$alpha_{1}=g_{1}^{*}, alpha_{2}=g_{2}^{*}, alpha_{3}=g_{3}^{*}, alpha_{4}=g_{4}^{*}$$
3) $f^{*}=beta _{1} g_{1}^{*}+beta _{2} g_{2}^{*}+beta _{3} g_{3}^{*}+beta _{4} g_{4}^{*}$ so I substitute the early calculated values.
linear-algebra
linear-algebra
asked Feb 2 at 21:50
MP3129MP3129
872211
872211
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