# 11.3 Give the matrix in standard basis

• MHB
• karush
In summary: HB, for your help. In summary, the matrix $\left[T\right]_\infty^\infty$ in the standard basis $\alpha=(x^2,x,1)$ is given by $\left[\begin{array}{c}1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0\end{array}\right]$ and in the second problem the ordered basis vectors are x^2+ x+ 1, x+1. and 1. The matrix representation of T in this ordered basis is $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 1 & 0 & 0\end{pmat karush Gold Member MHB We define the application$T:P_2\rightarrow P_2$by $$T(p)=(x^2+1)p''(x)-xp'(x)+2p'(x)$$ 1. Give the matrix$\displaystyle\left[T\right]_\infty^\infty$in the standard basis$\alpha=(x^2,x,1)$2 Give the matrix$\displaystyle\left[T\right]_\infty^\infty$where$\beta=\{x^2+x+1,x+1,1\}$would this be$\left[\begin{array}{c}1 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right]\$

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Do you see that $$\begin{pmatrix}a & b & c \\ d & e & f \\ g & h & i\end{pmatrix}\begin{pmatrix}1 \\ 0 \\ 0 \end{pmatrix}= \begin{pmatrix}a \\ d \\ g\end{pmatrix}$$,
that $$\begin{pmatrix}a & b & c \\ d & e & f \\ g & h & i\end{pmatrix}\begin{pmatrix}0 \\ 1 \\ 0 \end{pmatrix}= \begin{pmatrix}b \\ e \\ h\end{pmatrix}$$,
and $$\begin{pmatrix}a & b & c \\ d & e & f \\ g & h & i\end{pmatrix}\begin{pmatrix}0 \\ 0 \\ 1 \end{pmatrix}= \begin{pmatrix}c \\ f \\ i\end{pmatrix}$$?

That is, applying the linear transformation to the vectors in the ordered basis, in turn, gives you the columns of the matrix.

Here the linear transformation is $$T(p)= (x^2+ 1)p''(x)- xp'(x)+ 2p(x)$$ (you have "2p'(x) but that looks suspicious. If that were correct why wouldn't it be "(2- x)p'(x)"?) and, in the first problem, the basis vectors are $$x^2$$,$$x$$, and $$1$$. $$T(x^2)= (x^2+ 1)(2)- x(2x)+ 2x^2=x^2+ 1$$. That would be represented by the column matrix $$\begin{pmatrix}1 \\ 0 \\ 1\end{pmatrix}$$ and that is the first column of the matrix. $$T(x)=(x^2+ 1)(0)- x(1)+ 2(x)= x$$ which is represented by the column matrix $$\begin{pmatrix}0 \\ 2 \\ 0\end{pmatrix}$$ and that is the second column of the matrix. Finally, $$T(1)= (x^2+ 1)(0)- x(0)+ 2(1)= 2$$ which is represented by the column matrix $$\begin{pmatrix}0 \\ 0 \\ 2\end{pmatrix}$$ and that is the third column of the matrix. The matrix representation of T in this ordered basis is $$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 1 & 0 & 2 \end{pmatrix}$$.

In the second problem the ordered basis vectors are $$x^2+ x+ 1$$, $$x+1$$. and $$1$$. $$T(x^2+ x+ 1)= (x^2+ 1)(2)- x(2x+ 1)+ 2(x^2+ x+ 1)= 2x^2+ 2- 2x^2- x+ 2x^2+ 2x+ 2= 2x^2+ x+ 4$$. But now we have to write that in terms of this basis. That is, we need to find a, b, and c so that $$a(x^2+ x+ 1)+ b(x+ 1)+ c(1)= ax^2+ (a+ b)x+ (a+ b+ c)= 2x^2+ x+ 4$$. We have a= 2, a+ b= 2+ b= 1 so b= -1 and a+ b+ c= 2- 1+ c= 1+ c= 4 so c= 3. The first column of the matrix is $$\begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}$$. $$T(x+ 1)= (x^2+ 1)(0)- x(1)+ 2(x+ 1)= x+ 1$$. Well that's easy! $$x+ 1= 0(x^2+ x+ 1)+ 1(x+ 1)+ 0(1)$$ so the second column of the matrix is $$\begin{pmatrix}0 \\ 1 \\ 0\end{pmatrix}$$. $$T(1)= (x^2+ 1)(0)- x(0)+ 2(1)= 2$$. That can be written $$2= 0(x^2+ x+ 1)+ 0(x+ 1)+ 2$$ which corresponds to the column matrix $$\begin{pmatrix}0 \\ 0 \\ 2\end{pmatrix}$$. The matrix representing T in this ordered matrix is $$\begin{pmatrix} 2 & 0 & 0 \\ -1 & 1 & 0 \\ 3 & 0 & 2\end{pmatrix}$$.

I deeply appreciate the extended explanation
that helped quite a bit...

examples from books sometimes assume way too much

that is where MHB has filled in the best

Mahalo

## 1. What is a matrix in standard basis?

A matrix in standard basis is a representation of a linear transformation or a set of linear equations using a specific basis, typically the standard basis. The standard basis is a set of vectors that form the coordinate axes in a vector space.

## 2. How do you give a matrix in standard basis?

To give a matrix in standard basis, you first need to determine the vectors that form the standard basis. These are usually the unit vectors (e.g. [1,0,0], [0,1,0], [0,0,1] in 3-dimensional space). Then, you can represent the linear transformation or set of equations using these vectors as the columns of the matrix.

## 3. What is the purpose of using a standard basis for matrices?

The standard basis allows for a standardized representation of linear transformations or equations, making it easier to compare and analyze different matrices. It also simplifies calculations and allows for easier visualization of the transformation or solution.

## 4. How is a matrix in standard basis different from a matrix in another basis?

A matrix in standard basis is specifically represented using the standard basis vectors, while a matrix in another basis may use different vectors as the basis. This can result in different representations of the same transformation or equations.

## 5. Can a matrix in standard basis be converted to another basis?

Yes, a matrix in standard basis can be converted to another basis using a change of basis matrix. This involves finding the transformation matrix that maps the standard basis vectors to the new basis vectors, and multiplying it with the original matrix to get the new representation.

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