# Find the eingvalues of a Linear Transformation

Let $$V$$ be a real vector space and $$\{v_1,v_2, \cdots, v_n\}$$ a basis of $$V$$. If $$T:V \mapsto V$$ is a linear transformation such that $$T(v_1)=T(v_n)=v_1+v_n$$ and $$T(v_i)=v_i$$ for all $$i \in \{2,3, \cdots, n-1\}$$, prove that the eingvalues of $$T$$ are 0,1,2.

I can show that $$T(v_1 - v_n)=0$$. Does it proves that $$0$$ is an eingvalue?

In the same approach we can show that $$T(v_1 + v_n)=2(v_1 + v_n)$$ but im not sure if this answer the question or i have to express the linear transformation in term of a matrix, so far i found that $$T$$ can be written as

$$A=\begin{pmatrix} 1& 0& \cdots& 0& 1\\ 0& 1& 0& 0&0 \\ \vdots& 0 & 1 & 0& \vdots\\ 0& 0& 0&1 & 0\\ 1& 0& \cdots& 0& 1 \end{pmatrix}$$

but dont know how to finish from here.

• It is sufficient to show that $T(v_1 - v_n) = 0$ and $T(v_1 + v_n) = 2(v_1 + v_n)$. These are the definitions of eigenvalues and eigenvectors. – xbh Jan 13 at 3:58
• Note that this only shows that 0 and 2 are eigenvalues. You should also argue that 1 is an eigenvalue and that there are no other eigenvalues. – tch Jan 13 at 4:05
• got it, thanks. – ipreferpi Jan 13 at 4:23

Recall that a scalar $$\lambda$$ is called an eigenvalue of a linear transformation $$T:V\rightarrow V$$ if there is a nonzero vector $$v\in V$$, called an eigenvector of T, such that $$T(v)=\lambda v.$$ So the answer to your first question

I can show that $$T(v_1 - v_n)=0$$. Does it proves that $$0$$ is an eingvalue?

is "yes". For the vector $$v_1 - v_n$$ is nonzero. Similarly, the equations $$T(v_1+v_n)=v_1+v_n$$ $$T(v_i)=v_i(i=2,\dots,n-1)$$ imply that $$1$$ and $$2$$ are eigenvalues of $$T$$.

But how can we be sure that T has no other eigenvalues? Set $$v_{i}'=\begin{cases} v_{1}-v_{n} & \text{if }i=1\\ v_{i} & \text{if }i=2,\dots,n-1\\ v_{1}+v_{n} & \text{if }i=n \end{cases}.$$ The set of vectors $$\{v'_1, \dots, v'_n\}$$ will form a basis of $$V$$, with respect to which $$T$$ has the matrix representation $$A'=\left[\begin{array}{ccccc} 0\\ & 1\\ & & \ddots\\ & & & 1\\ & & & & 2 \end{array}\right].$$ Suppose that $$\lambda$$ is an eigenvalue of $$T$$, that $$v\in V$$ is an eigenvector corresponding to $$\lambda$$, and that $$x\in \mathbb{R} ^n$$ is the coordinate vector of $$v$$ with respect to the basis $$\{v'_1, \dots, v'_n\}$$, (i.e. if $$x=(x_1,\dots ,x_n)^T$$, then $$v'=x_1v'_1+\cdots +x_nv'_n)$$. Then we have $$A'x=\lambda x$$. We can now turn to direct computation to see that $$\lambda =0,1,$$ or $$2$$.

But in this case, it might actually be quicker to just systematically compute the characteristic polynomial of $$T$$, for $$\det(tI-A)$$ ($$A$$ is the matrix that you found) can be computed fairly easily using row and column expansion.

• it may be just a typo but its $det (A-tI)$. – ipreferpi Jan 13 at 7:19
• i did another approach since i didn't like to compute any determinant: since we have to solve $Tu=\lambda u$, where $u=\alpha_1v_1+\cdots,\alpha_nv_n$ it follows that we have to solve \\ - $\alpha_1+\alpha_n=\alpha_1\lambda$.\\ - $\alpha_i=\alpha_i\lambda$ \\ -$\alpha_1+\alpha_n=\alpha_n\lambda$. hence $\lambda=0,1,2$. – ipreferpi Jan 13 at 8:49
• corrected comment: i did another approach since i didn't like to compute any determinant: since we have to solve $Tu=\lambda u$, where $u=\alpha_1v_1+\cdots,\alpha_nv_n$ it follows that we have to solve (1) $\alpha_1+\alpha_n=\alpha_1\lambda$ . (2)$\alpha_i=\alpha_i\lambda$ . (3)$\alpha_1+\alpha_n=\alpha_n\lambda$. hence $\lambda=0,1,2$. – ipreferpi Jan 13 at 9:00
• Thank you for pointing out the typo. Your alternative approach works fine, as long as your logic between the sentence "it follows that we have to solve" and the next "hence" is correct. – Ken Jan 13 at 10:43