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My quation is about Linear transformations.

My question is: given 2 different matrices $A, B$ such that:

$$A\cdot ( 1, 0, 1)= B\cdot (1, 0, 1)$$

$$A\cdot ( 1, 1, 1)= B\cdot (1, 1, 1)$$

So, let's define a linear transformation $T:R^3 \to R^3$ such that:

$$\forall \overrightarrow v \in R^3 \Rightarrow T(\overrightarrow v)=(A-B)\cdot \overrightarrow v$$

  • Find $dim(Ker(T)).$

  • Find $dim(Im(T)).$

  • Given that $T(0, 0, 1)=(1, 0, 1),$ Find $[T],$ $\bigl( [T] \cdot \overrightarrow v = T(\overrightarrow v), [T]$ is the matrix representation of $T$ $\bigr).$

My attempts:

  • $dim(Ker(T)) = ?$

Since: $$A\cdot ( 1, 0, 1)= B\cdot (1, 0, 1)$$

$$A\cdot ( 1, 1, 1)= B\cdot (1, 1, 1)$$

And the matrix representation is: $[T] = (A-B),$ I conclude that: $T(1, 0, 1)=T(1, 1, 1) = (0, 0, 0),$ And these 2 vectors are linearly independent in $R^3$ so $dim(Ker(T)) = 2.$

  • $dim(Im(T)) = ?$

Since $dim(Ker(T))= 2,$ and $dim(Ker(T)) + dim(Im(T)) = 3,$ We get that $dim(Im(T)) = 1.$

  • Given that $T(0, 0, 1)=(1, 0, 1),$ $[T] = (A - B) = ?$

I put the 3 vectors as columns in the left side of the matrix and their transformation in the right side, and by Gaussian elimination i get the identity matrix $I$ on the left side, and the results that i get on the right side (which are supposed to be the transformations of the 3 unit vectors in the standard basis of $R^3$) are not correct.

I know that the columns of $[T]$ are the transformations of the 3 unit vectors in the standard basis of $R^3$, but how to find these transformations?

I don't know how to solve that problem, any suggestions?

Thanks for help!!

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Yes since the matrices are different and $T(1, 0, 1)=T(1, 1, 1) = (0, 0, 0)$ we can conclude that

  • $\dim(\ker(T)) = 2$
  • $\dim(\operatorname{Im}(T)) = 1$

To find $[T]$ we have all the information since we know how $T$ acts on a basis, indeed

$$[T]\begin{bmatrix}1&1&0\\0&1&0\\1&1&1\end{bmatrix}=\begin{bmatrix}0&0&1\\0&0&0\\0&0&1\end{bmatrix}\implies [T]=\begin{bmatrix}0&0&1\\0&0&0\\0&0&1\end{bmatrix}\begin{bmatrix}1&1&0\\0&1&0\\1&1&1\end{bmatrix}^{-1}$$

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