# Linear algebra Please help me to solve this problem .Is $R(T)=R(T^2)$?

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 ?

• Can you write out explicitly the formula for $T^2$? – Arthur Dec 1 '16 at 9:40
• Here T^2 means composition ToT – Heet Modi Dec 1 '16 at 9:48
• What I meant was, have you tried writing out a formula for $T^2$, using coordinates? What is $T^2(x_1, x_2, x_3)$? – Arthur Dec 1 '16 at 9:49
• T^2 =(14x1+17x2+3x3,17x1+26x2+9x3,3x1+9x2+6x3) – Heet Modi Dec 1 '16 at 9:58
• 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? – Arthur Dec 1 '16 at 10:02

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 $$T^2=A^2$$
• 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. – Arthur Dec 1 '16 at 10:04