• 1.) Show that any nonzero linear combination of two eigenvectors v,w corresponging to the same eigenvalue is also an eigenvector.

  • 2.) Prove that a linear combination $cv+dw$, with $c,d \ne 0$, of two eigenvectors corresponding to different eigenvalues is never an eigenvector.

  • 3.) Let $\lambda$ be a real eigenvalue of the real n x n matrix A, and $v_1,...,v_k$ a basis for the associated eigenspace $V_{\lambda}$. Suppose $w \in \mathbb{C^n}$ is a complex eigenvector, so $Aw = \lambda w$. Prove that $w = c_1v_1 + ... + c_kv_k$ is a complex linear combination of the real eigenspace basis.

For 1 and 2 I know that if two eigenvectors $\vec{v}_1$ and $\vec{v}_2$ are associated with the same eigenvalue then any linear combination of those two is also an eigenvector associated with that same eigenvalue. But, if two eigenvectors $\vec{v}_1$ and $\vec{v}_2$ are associated with different eigenvalues then the sum $\vec{v}_1+\vec{v}_2$ need not be related to the eigenvalue of either one. In fact, just the opposite. If the eigenvalues are different then the eigenvectors are not linearly related. But I can show this using a proof?

For 3. I am not too sure.


For 2. let $A$ be the matrix; calculate $A(cv+dw)$ under the assumption that $v$ and $w$ are eigenvectors belonging to different eigenvectors; see whether the result is consistent with $cv+dw$ being an eigenvector.

For 3. write $w=w_1+iw_2$ where $w_1$ and $w_2$ are real, and then see what $Aw=\lambda w$ tells you about $w_1$ and $w_2$.

EDIT: In view of the length of the discussion in the comments, I'll expand on this last part.

We get $$A(w_1+iw_2)=\lambda(w_1+iw_2)$$ which is $$Aw_1+iAw_2=\lambda w_1+i\lambda w_2$$ Now if $a+bi=c+di$ where $a,b,c,d$ are real (real numbers, or vector with real entries, or matrices with real entries), then necessarily $a=c$ and $b=d$ --- that's what equality means in the complex realm. So we deduce $$Aw_1=\lambda w_1,\qquad Aw_2=\lambda w_2$$ So $w_1$ and $w_2$ are in the eigenspace $V_{\lambda}$. We are told $v_1,\dots,v_k$ is a basis for $V_{\lambda}$, so $$w_1=r_1v_1+\cdots+r_kv_k,\qquad w_2=s_1v_1+\cdots+s_kv_k$$ for some real numbers $r_1,\dots,r_k,s_1,\dots,s_k$. Then $$w=w_1+iw_2=c_1v_1+\cdots+c_kv_k{\rm\ with\ }c_j=r_j+is_j,j=1,\dots,k$$

  • $\begingroup$ Thanks a lot! Will I use the same method you used for part 2 for part 1, but instead showing that it is consistent instead of inconsistent? $\endgroup$ – Q.matin Nov 20 '12 at 3:18
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    $\begingroup$ Yes (I didn't mention part 1 in my answer because I misunderstood you to be saying you had already done that part). $\endgroup$ – Gerry Myerson Nov 20 '12 at 4:42
  • $\begingroup$ I am having trouble proving part3. Do you think you can add more detail please? What do they mean when they say that $v_1,...,v_k$ are the basis for the eigenspace? $\endgroup$ – Q.matin Nov 20 '12 at 23:42
  • $\begingroup$ The eigenspace $V_{\lambda}$ is the set of all vectors $v$ such that $Av=\lambda v$. You can prove that $V_{\lambda}$ is a vector space, a subspace of ${\bf R}^n$. Being a (finite-dimensional) vector space, it has a basis. They are saying, let $v_1,\dots,v_k$ be a basis for the vector space $V_{\lambda}$. $\endgroup$ – Gerry Myerson Nov 21 '12 at 0:50
  • $\begingroup$ so far what I understand is that $v_1,...,v_k$ is basis of the real eigenvalue. So I have to prove that the linear combination of the basis equals w which is a complex eigenvector. Is that right? If it is, how does the basis of the real eigenvalue, $v_1,...,v_k$ make the linear combinations of it equal a complex eigenvector. I am having trouble understanding that. $\endgroup$ – Q.matin Nov 21 '12 at 1:52

All eigenvectors of a matrix are of the form $\lambda v$ where lambda is a scalar and v is any one of the eigenvalues.

$v+w=\lambda v \implies w=(\lambda-1)v$

The eigenspaace is the space of al eigenvectors with a given eigenvalue so I guess that the question meant to ask for the eigensystem. The definition of a basis is a set of vectors that all the eigenvectors are linear combinations of, so there doesn't seem to be anything to prove fr part 3.

  • $\begingroup$ Thanks and I am not sure either? Does that also follow question 1 and 2? $\endgroup$ – Q.matin Nov 20 '12 at 2:56

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