$T:\mathbb{R}^3\to \mathbb{R}^3$ is defined by $T(x,y,z)=(x+y,y+z,z-x)$
Then I need too find an orthonormal basis for Range of $T$
$T(1,0,0)=(1,0,-1),T(0,1,0)=(1,1,0),T(0,0,1)=(0,1,1)$ could any one help me the steps? Thank you.
[Math] o find an orthonormal basis for Range of $T$
linear algebra
Related Solutions
Here it is very easy to see how the standard basis vectors can be written as linear combinations of the vectors in B.
Example:
$(1,0,0)=(1,1,0)-(0,1,0)$
Therefore $T((1,0,0))=T((1,1,0))-T((0,1,0))$
Moreover we know that $T((1,1,0))=2(1,1,0)+n(0,1,0)=(2,2+n,0)$ and that $T((0,1,0))=(m,m,0)$, whence $T((1,0,0))=(2-m,2+n-m,0)$.
This is the first column in your sought after matrix, and you can do similarly with the other two standard basis vectors.
The more general approach is to first find the change of basis matrices $P_{\mathcal{B}\rightarrow{\mathcal{E}}}$ and $P_{\mathcal{E}\rightarrow{\mathcal{B}}}$, where $\mathcal{E}$ denotes the standard basis. One then has the following formula:
$[T]_{\mathcal{E}}=P_{\mathcal{E}\rightarrow{\mathcal{B}}}[T]_{\mathcal{B}}P_{\mathcal{B}\rightarrow{\mathcal{E}}}$
Good work so far. To complete the problem, the key is to understand the meaning of the matrix of a linear map. Consider a linear map $T:V \rightarrow W$. For simplicity, assume the basis for $V$ is $\{\alpha_{1},\alpha_{2},\alpha_{3}\}$ and the basis for $W$ is $\{\beta_{1},\beta_{2}\}$. The first step, as you did above, is to consider the action of $T$ on the basis vectors in $V$. Say we did that and we have $T(\alpha_1), T(\alpha_2), T(\alpha_3)$. Now, we need to express each of these vectors that lie in $W$ as a linear combination of the basis in $W$ i.e $\{\beta_{1},\beta_{2}\}$ \begin{align} T(\alpha_1) &= a_{11} \beta_{1} + a_{12} \beta_{2} \\ T(\alpha_2) &= a_{21} \beta_{1} + a_{22} \beta_{2} \\ T(\alpha_3) &= a_{31} \beta_{1} + a_{32} \beta_{2} \end{align} where the $a_{ij}$ are some coefficients you find. As an example, the coordinate representation of $T(\alpha_1)$ is simply $$ T(\alpha_1)_{\beta}= \begin{pmatrix} a_{11}\\ a_{12} \end{pmatrix} $$ where the subscript $\beta$ denotes the coordinate represetation with respect to the basis $\beta$. The matrix is nothing but the coordinate representation of the mapped vectors i.e $$ \begin{pmatrix} \vdots & \vdots & \vdots \\ T(\alpha_1)_{\beta} & T(\alpha_2)_{\beta} & T(\alpha_3)_{\beta}\\ \vdots & \vdots & \vdots \\ \end{pmatrix} $$ As you can see, its representation depends on the basis we choose. The operator is one and one but based on the basis you choose, you have different matrix representations. The matrix of the linear map is then $$ \begin{pmatrix} a_{11} & a_{21} & a_{31}\\ a_{12} & a_{22} & a_{32} \end{pmatrix} $$ So for your problem, all that remains to be done is to compute the coordinate representation of the mapped vectors interms of the basis in $\mathbb R^{2}$.
Best Answer
First we see that the vectors are dependent (take the determinant of the matrix that has them as rows and you will get $0$). Note that $(1,1,0)$ and $(0,1,1)$ are independent, so they form a basis. However, this is not an orthonormal basis. The use Gram-Schmidt:
Let
$u_1=(1,1,0)$, and
$u_2=(0,1,1)-\frac{(1,1,0)\cdot(0,1,1)}{(1,1,0)\cdot(1,1,0)}(1,1,0)$ $=(0,1,1)-\frac{1}{2}(1,1,0)=(-1/2,1/2,1)$
This is an orthogonal basis. Divide each one by their norm to obtain an orthonormal basis.