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The options given are :

a. Linear dependence/ independence cannot be determined without knowing the whole matrix A.

b. The vectors are linearly dependent

c. The vectors are linearly independent iff A is symmetric

d. The vectors are linearly independent iff A is PD (Positive Definite)

e. The vectors are linearly independent

For background, I tried to apply the principle of Krylov subspaces here and ended up with a solution that the vectors are almost always linearly independent, albeit with some exceptions. Some exceptions among diagonalizable matrices occur when A has an eigenvalue of 0,1, or a root of unity and x is an associated eigenvector.

So I am now confused as to which option to select. Any help appreciated.

Adithya52
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2 Answers2

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Item $(e)$ is false by taking $x=0$. This would make $\{Ax,\ldots ,A^5x\}=\{0\}$ which is clearly linearly dependent. To see why items $(b)-(d)$ fail, take $x=e_1$, and consider the $5\times 5$ matrix $A$ defined by $$A=\begin{pmatrix}0&0&0&0&-1\\ 1&0&0&0&0\\ 0&1&0&0&0\\ 0&0&1&0&0\\ 0&0&0&1&0\end{pmatrix}$$ It is easily verified that $\{Ax,A^2x,\ldots,A^5x\}$ is linearly independent. However, $A$ is not symmetric nor positive definite as $A$ has an eigenvalue of $-1$.

Matthew H.
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The statement that is the least false is a.

A matrix $A$ will be such that there exists an $x$ for which $\{Ax,\dots,A^5 x\}$ is linearly independent if and only if $A$ is invertible and non-derogatory. For such a matrix $A$, a vector $x$ will have this property if and only if it is an "$A$-cyclic vector", i.e. if the set $\{x,Ax,A^2x,\dots\}$ spans $\Bbb R^5$.

Statement b is false because the vectors are independent, for instance, when $A$ an invertible diagonal matrix and $x = (1,\dots,1)^T$. Statement c is false: the vectors are not independent if $A$ is the identity matrix. Statements d and e are false for the same reason.

Ben Grossmann
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