Take the 2-minute tour ×

I am tring to solve the following task:

Linear transformation $f: \mathbb R^2 \rightarrow \mathbb R^2$ is given by $f(\begin{bmatrix} x_1\\ x_2 \end{bmatrix}) = \begin{bmatrix} 2x_1-x_2\\ x_1+x_2 \end{bmatrix}$. Answer true or false to the following questions:

a) in some basis of transformation $ \mathbb R^2$ transformation matrix of $f$ is $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

b) $f$ is a bijection

c) transformation matrix of $f$ in basis $([1,0]^T, [1,1]^T)$ is $\begin{bmatrix} 1 & 1 \\ -1 & 2 \end{bmatrix}$

Could you kindly give me any HINTS how to start it (not solution)?

share|improve this question
a) How are the matrices of a linear transformation in different bases related. b) What criterion have you for a linear transformation to be a bijection? c) Calculate $f((1,0))$ and $f((1,1))$ and see whether the results match. –  Daniel Fischer Jun 28 '13 at 14:02

1 Answer 1

up vote 2 down vote accepted

Hints (assuming some familiarity with various results of linear algebra):

a) If $A$ and $B$ are transformation matrices of $f$ with respect to different bases, then there is some matrix $X$ such that $XAX^{-1} = B$. What does this say if $A$ is the identity matrix?

b) A linear transformation is bijective if and only if for any (and thus all) basis, its transformation matrix is invertible.

c) This is a calculation.

share|improve this answer
+1 Ok, thanks a lot for your help! I have obtained False, True, False. But b) point I have proved not by criterion with basis, but checking conditions on suriection and injection. Could you explain me how to prove it using your criterion (with basis and invertible matrix)? –  JosephConrad Jun 28 '13 at 15:28
Do you believe that the criterion is true? If so, in the standard basis, the matrix for $f$ is $\pmatrix{ 2 & -1 \\ 1 & 1 }$. Now, there are tons of criteria for a matrix being invertible, typically the easiest one to check being that the matrix has non-zero determinant, which is definitely the case for the matrix here. –  fuglede Jun 28 '13 at 21:32

Your Answer


