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I understand that if a matrix is singular, it has no inverse. If it has nontrivial solutions, it means at least one solutions exists. If it is linearly dependent, it means that for $a_1 \mathbf{v_1}+a_2 \mathbf{v_2} + ... + a_n \mathbf{ v_n} =\mathbf{ 0}$. Not all the $a$'s are 0. (Not all the coefficients of v_k are zero to satisfy the equation. How does singular relate to nontrivial solutions and nontrivial solutions relate to linear dependent?

Let's say you have 3 vectors: $$\vec p_1(x)=a_1x^2+b_1x+3$$ $$\vec p_2(x)=a_2x^2+b_2x+4$$ $$\vec p_3(x)=a_3x^2+b_3x+99$$

We multiply all the stuff with c and get $$c_1\vec p_1(x)+c_2\vec p_2(x)+c_3\vec p_3(x)=0$$

Then, we make $$a_1c_1+a_2c_2+a_3c_3=0$$ $$b_1c_1+b_2c_2+b_3c_3=0$$ $$3c_1+4c_2+99c_3=0$$

This coefficient matrix can be singular hence there are nontrivial solutions. So, $\vec p_1$, $\vec p_2$ and $\vec p_3$ are linearly dependent.

OR

This coefficient matrix can be nonsingular hence there are trivial solutions. So, $\vec p_1$, $\vec p_2$ and $\vec p_3$ are linearly independent.

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1 Answer 1

up vote 3 down vote accepted

Let me answer your question for square matrices.

Let $A$ be an $n\times n$ matrix. Then the following are equivalent:

  1. $A$ is invertible.
  2. $A$ is row equivalent to the identity.
  3. $A$ rank $n$.
  4. $A$ has a trivial kernel.
  5. $A$ has linearly independent columns.
  6. $A$ has linearly independent rows.

This is often known as (a part of) the Invertible Matrix Theorem.

If you have a set of vectors expressed in coefficients with respect to some basis then your vectors will be linearly independent if and only if the resulting matrix is invertible. For your example, suppose that we work in $P_2$, the space of polynomials degree at most $2$. Let us write your vectors with respect to the standard basis $\mathcal{B} = \{1,\ x,\ x^2\}$. Then $$[p_1]_\mathcal{B} = \begin{pmatrix}3 \\ b_1 \\ a_1\end{pmatrix},\ \ \ [p_2]_\mathcal{B} = \begin{pmatrix}4 \\ b_2 \\ a_2\end{pmatrix},\ \ \ [p_3]_\mathcal{B} = \begin{pmatrix}99 \\ b_3 \\ a_3\end{pmatrix}$$ If we form the matrix with these vectors as our columns $$A = \begin{pmatrix} 3 & 4 & 99 \\ b_1 & b_2 & b_3 \\ a_1 & a_2 & a_3\end{pmatrix}$$ Then our vectors will be linearly independent if and only if this matrix is non-singular.

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What's a trivial kernel? –  raindrop Apr 1 '13 at 16:04
    
$\left\{0_E\right\}$ –  xavierm02 Apr 1 '13 at 16:25
    
Here's an explanation of points 1-4 of the Invertible Matrix Theorem provided in the answer: 1. $A^{-1}$ exists. 2. $A$ can be obtained from the identity matrix, $I$ by a finite number of row operations. 3. All columns and rows of $A$ are linearly independent. 4. "For a linear map $f:A \to B$, the kernel of $f$ is the set of elements of $A$ that map to $0$ in $B$. The kernel is trivial if it contains only the single element $0$ (which must map to $0$ in $B$ by linearity)." - Henning Makholm chat.stackexchange.com/transcript/message/8765408#8765408 –  raindrop Apr 1 '13 at 20:18

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