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Let's say if I defined a $3\times 2$ matrix $B$: \begin{bmatrix} 1 & 0\\ 0 & 1\\ 0 & 0\end{bmatrix} and said that if my system is homogeneous $Bx = 0$ then the column vectors of $B$ must be linearly independent.

My column vectors are $b_1 = \begin{bmatrix} 1\\ 0 \\0 \end{bmatrix}$ and $b_2 = \begin{bmatrix} 0 \\1 \\0 \end{bmatrix}$.

My solution would be $x_1= 0$ and $x_2=0$?

But I also have the $0 = 0$ condition at the last row. Does that mean that my system hast infinitely many solutions that include the zero solution? Therefore my colum vectors are not linearly independent ? Therefore if my column vectors were to be a set of eigenvectors since they are linearly dependent then the matrix (not $B$) from which I found my eigenvalues (and consequently my eigenvectors) is not diagonizable? Is my procedure to test whether a set of vectors is linearly independent adequate? Would appreciate your insight! Thanks!

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We say a square matrix, $A$ is diagonalizable if there exists invertible matrix $P$ and a diagonal matrix $D$ such that $A=PDP^{-1}$. Matrix $B$ is not a square matrix here.

In solving the system $Bx=0$, the unique solution would be $x=0$. The constraints that the system has to satisfy are $x_1=0$, $x_2=0$, $0=0$. The third constraint is trivially true and $(x_1,x_2)=(0,0)$ is the only solution due to the first two constraint. Notice that matrix $B$ is already in an RREF form. A homogeneous system has infinitely many solution if there is a non-pivot column, which is not the case here.

The columns are independent.

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  • $\begingroup$ Thanks! I see my confusion now! Yes, they are linearly independent. $\endgroup$ – dareToDiffer07 Aug 8 '17 at 21:31

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