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Putting all vectors (matrices) in one gives

$$ \begin{bmatrix} 1 & 0 & 0 & m_1 & \cdots\\ 0 & 1 & 0 & m_2 & \cdots\\ 0 & 0 & 1 & m_3 & \cdots\\ \end{bmatrix}$$

$rref$ of this matrix is always that matrix.

Does this mean that the dimension of $span\{I,M,M^2,...\}$ is $3$ and a basis is formed of column vectors of identity matrix?

But because $I$ can't be represented as a linear combination of, for example, $M$ and $M^2$, from here it follows that a basis for $span\{I,M,M^2,...\}$ is an empty set and the dimension is $0$.

Which is correct?

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  • $\begingroup$ 1: The matrix you give is not square. 2: A basis of your $\text{span}$ will consist of matrices, not of (column) vectors. 3: I can't make sense of your last sentence about representing $I$. Would you care to expand? $\endgroup$ – Servaes Feb 9 '16 at 13:35
  • $\begingroup$ What is the matrix $M$? $\endgroup$ – Bernard Feb 9 '16 at 13:44
  • $\begingroup$ @Bernard $M$ is invertible squared matrix over $\mathbb{R}$. $\endgroup$ – user300048 Feb 9 '16 at 13:52
  • $\begingroup$ I understood that. I mean ‘What is $M$ w.r.t. the displayed matrix in the second line of your post?’ $\endgroup$ – Bernard Feb 9 '16 at 13:54
  • $\begingroup$ @Bernard A matrix that contains of all matrices from $span$ set. Finding $rref$ of that matrix would result in knowing the set of linearly independent vector that form a basis. Note that I arbitrarily chose $n=3$ is the order of matrices in $span$ set. $\endgroup$ – user300048 Feb 9 '16 at 14:07
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Dimension has to be the degree of the minimal polynomial of M

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  • $\begingroup$ Perhaps it would improve the answer to explain why. $\endgroup$ – Travis Willse Feb 9 '16 at 13:15
  • $\begingroup$ Because the minimal polynomial gives u relation between the powers of M and that can be used to eliminate the dependent powers..ex if the minimal polynomial is of degree m say then u get M^m=linear combination of I,M.......M^m-1.hence the dimension m..i hope it helps $\endgroup$ – user311790 Feb 9 '16 at 13:20
  • $\begingroup$ My point is that, for someone who knows about the minimal polynomial but doesn't see the connection with the question, this answer is not likely to be very illuminating without some more details. $\endgroup$ – Travis Willse Feb 9 '16 at 13:22
  • $\begingroup$ Could you enlighten mein with that "for someone"of yours $\endgroup$ – user311790 Feb 9 '16 at 13:23
  • $\begingroup$ I'm not quite sure what you mean? $\endgroup$ – Travis Willse Feb 9 '16 at 13:24
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Every square matrix is a zero of its own minimal polynomial. So plugging $M$ into its minimal polynomial gives a linear combination of powers of $M$ which is equal to zero, making these powers of $M$ linearly dependent. This shows that the dimension of your space is at most the degree of the minimal polynomial. By minimality of the degree of the minimal polynomial, the dimension of your space is exactly the degree of the minimal polynomial.

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  • $\begingroup$ How to find a basis for $span\{I,M,...\}$? If $\dim(span)=n$, then how to find $n$ matrices from $span$ to form a basis? $\endgroup$ – user300048 Feb 9 '16 at 13:59
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    $\begingroup$ @user300044 It turns out, in your particular case, that the matrices $\{I,M,\dots,M^n\}$ can always be taken as your basis (if the dimension of the span is $n$). $\endgroup$ – Ben Grossmann Feb 9 '16 at 14:32
  • $\begingroup$ @Omnomnom I think you mean $\{I,M,\ldots,M^{n-1}\}$ is a basis if the dimension of the span is $n$? $\endgroup$ – Servaes Feb 10 '16 at 13:57
  • $\begingroup$ Yes, that's what I mean $\endgroup$ – Ben Grossmann Feb 10 '16 at 14:34

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