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Let $T:\mathbb{R}^3\to\mathbb{R}^3$ be a L.T. defined by $$T(x_1,x_2,x_3)=(x_1+3x_2+2x_3\,,3x_1+4x_2+x_3\,,2x_1+x_2-x_3)$$ then the dimension of the range space of $T^2$ is ?

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  • $\begingroup$ Can you write out explicitly the formula for $T^2$? $\endgroup$ – Arthur Dec 1 '16 at 9:40
  • $\begingroup$ Here T^2 means composition ToT $\endgroup$ – Heet Modi Dec 1 '16 at 9:48
  • $\begingroup$ What I meant was, have you tried writing out a formula for $T^2$, using coordinates? What is $T^2(x_1, x_2, x_3)$? $\endgroup$ – Arthur Dec 1 '16 at 9:49
  • $\begingroup$ T^2 =(14x1+17x2+3x3,17x1+26x2+9x3,3x1+9x2+6x3) $\endgroup$ – Heet Modi Dec 1 '16 at 9:58
  • $\begingroup$ Good. Now you can examine the range space, i.e. the span of the three vectors $(14,17,3), (17, 26, 9)$ and $(3, 9, 6)$ (that is, $T^2(1, 0,0), T^2(0,1,0)$ and $T^2(0,0,1)$ respectively). Are they linearly independent? Is one a linear combination of the other two? Are all three of them multiples of one another? $\endgroup$ – Arthur Dec 1 '16 at 10:02
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Hint $$T(1,0,0)=(1,3,2)\\ T(0,1,0)=(3,4,1)\\ \quad T(0,0,1)=(2,1,-1)$$

we have $$T(x)=\underbrace{\left(\begin{matrix}1&3&2\\3&4&1\\2&1&-1\end{matrix}\right)}_{A}\left(\begin{matrix}x_1\\x_2\\x_3\end{matrix}\right)$$ thus $$T^2=A^2$$

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  • $\begingroup$ Thechnically, you don't have $T^2 = A^2$. $T$ is a linear transformation, and $A$ is a matrix representation of that linear transformation in some specific basis. But $A^2$ represents $T^2$ in that same basis (matrix multiplication is defined the way it is in order to make that work), and I'm just nit-picking. $\endgroup$ – Arthur Dec 1 '16 at 10:04

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