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- Summary
- Show that the linear transformation with the matrix below with respect to a basis for ##\textbf{R}^4## is a projection on a subspace ##U'## along a subspace ##U''##. Find ##U'##.

## \dfrac{1}{2}

\left(\begin{array}{rrrr}

0 & 2 & -2 & 0 \\

3 & -2 & 2 & -3 \\

3 & -4 & 4 & -3 \\

-2 & 2 & -2 & 2

\end{array}\right)##

\left(\begin{array}{rrrr}

0 & 2 & -2 & 0 \\

3 & -2 & 2 & -3 \\

3 & -4 & 4 & -3 \\

-2 & 2 & -2 & 2

\end{array}\right)##

The matrix satisfies ##A^2=A##, so it is a projection. To find ##U'##, one can find the ##\text{ker} \ (A-I)=\text{ker} \ (I-A)=\text{im} \ (A)=U'##. Also, ##\text{ker} \ (A-I)=\text{ker} \ (2(A-I))##.

Solving ##2(A-I)\textbf{x}=\textbf{0}## using Gauss Jordan elimination yields ##\textbf{x}=r(-2,-1,1,0)+t(-3,-3,0,1)## for ##t,r\in\textbf{R}##. But the solution given is ##\textbf{x}=r(0,3,3,-2)+t(1,-1,-2,1)##. Are these solutions equivalent? Since the columns of the matrix span the image, every vector in the image can be represented by at least the linear independent vectors in the matrix, which the latter solution does.