If a matrix is invertible, what does this tell us application wise?

I am familiar with what it implies in regards to the properties of the matrix, i.e: the determinant is non-zero, and for a matrix $A$, $Ax=0$ implies $x=0$.

However, during discussions in my lectures for an optimization class, a class mate always brings up non-invertible and invertible matrices and what they imply. I can't be too specific as I tend to get lost in discussion whenever this happens, but could anyone clue me in on what the significance of invertibility and linear independence (of which the definitions I am aware of) is?

Kind regards.


I had a lot of trouble getting my head around this myself and apologise if anything in my answer is incorrect.

Consider the matrix expression $$\begin{bmatrix} 3&6\\ 1&2 \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix}$$ I can take a factor of 3 out of the 1st row $$ 3\begin{bmatrix} 1&2\\ 1&2 \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix}$$ The matrix has a zero determinant by inspection or $2-2=0$

A matrix with zero determinant is singular and has no inverse.

Notice that the 1st row is obviously a linear combination of the second row and so they are linearly dependent. This was just an example to get a feeling for what is happening. It is more appropriate to think of the determinant as being designed to test for linear dependence. Herbert Gross has excellent lectures from MIT.

  • $\begingroup$ Thank you very much for this! This is what I was hoping for, and I'll definitely check out the MIT lectures. $\endgroup$ – yaweh Feb 1 '15 at 19:08

Hint : The Caley-hamilton-theorem states that a matrix with characteristic equation $$a_n\lambda^n+...+a_1\lambda+a_0=0$$ satisfies the equation $$a_nA^n+...+a_1A+a_0I=0$$

Also note, that A can be written as AI.

Now, do you see how $A^{-1}$ can be expressed by powers of $A$ , including I ?

  • $\begingroup$ Sorry, I'm not entirely sure what point you are trying to make. $A^{-1}(a_nA^n + ... + a_1A + a_0I) = a_nA^{(n-1)} + ... + a_1I + a_0A^{-1}$ ? $\endgroup$ – yaweh Feb 1 '15 at 17:43
  • $\begingroup$ First subtract $a_0I$, then the left side is of the form $AB$ with some matrix $B$. $\endgroup$ – Peter Feb 1 '15 at 17:44
  • $\begingroup$ If you divide by $-a_0$, you see the inverse. $\endgroup$ – Peter Feb 1 '15 at 17:46

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