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Let $L: \mathbb{R}^3 \to \mathbb{R}^3$ be a linear function defined by the formula $$ L\left( \begin{pmatrix}x\\y\\ z\end{pmatrix}\right) = \begin{pmatrix}x-3y \\ x+6y \\ 4x+6y\end{pmatrix} $$ Let $H = \operatorname{Im}(L)$ be the image of $L$. Let $$\beta = \left(\begin{pmatrix}1\\0\\2\end{pmatrix}, \begin{pmatrix}0\\1\\2\end{pmatrix}\right)$$ be a basis of $H$.

  1. Find the matrix $B$ that represents the restriction of the standard scalar product to the $H$ subspace (which is a plane), in respect to the basis $\beta$
  2. Find the matrix $A$ that represents the restriction $L' : H \to H$ of the linear function $L$, in respect to the basis $\beta$

MY PROBLEMS

My problem is with the concept of restriction: I don’t know how to obtain the restriction of a function to a subspace of its domain/codomain. I would be glad if you could explain the general way to do this type of exercises that involve restrictions, or if you could provide the theory behind calculating restrictions of linear maps and their representative matrices.

Also, I don’t know how to obtain the matrix that represents the restriction of a scalar product, or that represents the same scalar product but in respect to a different basis.

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1 Answer 1

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Restriction means that you will input only elements of $H$. You have a basis of $H$. The elements that you will input are linear combinations of that basis.

The two problems that you have are about computing matrices of linear (the $L$) and bilinear (the inner product). Matrices represent these by taking coordinates of the input in certain basis and giving you the output, or the coordinates of the output in that basis. So, if a vectors $x,y\in H$ have coordinates $\mathbf{x},\mathbf{y}$ in the basis $\beta$, then the inner product $x\cdot y$ can be computed as $\mathbf{x}^TB\mathbf{y}$. Likewise, the coordinates of $L(x)$ in the basis $\beta$ will be $A\mathbf{x}$, where $A$ is the matrix of $L$.

In the computations below I will use $[]$ to write coordinates in the basis $\beta$, and $()$ for the original $\mathbb{R}^3$ vectors.


To obtain $B$ we need to compute the products of the elements of $H$ that have coordinates $\begin{bmatrix}1\\0\end{bmatrix}$ and $\begin{bmatrix}0\\1\end{bmatrix}$ in the basis $\beta$. These are the elements of $\beta$ themselves. Let me call them $\beta_1$ and $\beta_2$, in the order that you wrote them.

Computing the "standard" inner products $\beta_1\cdot\beta_1=5$, $\beta_1\cdot\beta_2=4$, and $\beta_2\cdot\beta_2=5$.

Therefore, $$B=\begin{bmatrix}5&4/2\\4/2&5\end{bmatrix}$$


For the second part, you want to compute the coordinates of $L(\beta_1)$ and $L(\beta_2)$ in the basis $\beta$.

We have that $L(\beta_1)=\begin{pmatrix}1\\1\\4\end{pmatrix}=\beta_1+\beta_2$ and $L(\beta_2)=\begin{pmatrix}-3\\6\\6\end{pmatrix}=-3\beta_2+6\beta_2$

Therefore, the matrix of $L$ in the basis $\beta$ is $$\begin{bmatrix}1&-3\\1&6\end{bmatrix}$$

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