Matrix and Quadratic Formula

[piquagirl]

Can one use the quadratic formula for a 2x2 quadratic matrix?

 

[Ishuda]

Can one use the quadratic formula for a 2x2 quadratic matrix?

Do you mean solve the following for A?
a A² + b A + c = 0
where 0, a, b, and c are given scalars and A is a 2X2 matrix. The answer is no. In this particular case the quadratic equation would depend on the scalars a, b, and c and give a scalar answer when what you are looking for is a 2X2 matrix answer.

If this is not your question, would you give an example of what you are asking?

 

[piquagirl]

Do you mean solve the following for A?
a A² + b A + c = 0
where 0, a, b, and c are given scalars and A is a 2X2 matrix. The answer is no. In this particular case the quadratic equation would depend on the scalars a, b, and c and give a scalar answer when what you are looking for is a 2X2 matrix answer.

If this is not your question, would you give an example of what you are asking?

Thank you for responding Ishuda. I was thinking about a problem like X² - 5X + 6 = 0.

 

[stapel]

I was thinking about a problem like X² - 5X + 6 = 0.

Okay. And what are you wanting to do with a matrix, with respect to the given quadratic?

Please provide an example, along with a clear listing of your thoughts and efforts so far. Thank you!

 

[Ishuda]

Thank you for responding Ishuda. I was thinking about a problem like X² - 5X + 6 = 0.

If X is a scalar we have X² - 5X + 6 = (X - 2) (X - 3) = 0. However if X is a matrix, what we really mean by the factored form is (X - 2 I) (X - 3 I) where I is the identity matrix. So, lets, look at it

. . .\(\displaystyle (X\, -\, 2\, I)\, (X\, -\, 3\, I)\, =\, X^2\, -\, X\, 3\, I\, -\, 2\, I\, X\, -\, 2\, I\, \cdot\, (-3\, I)\)

Since the Identity matrix commutes with X and scalars commute with matrices, we can write this as X² - 5X + 6 I. Ah ha, you say, this does work (if I just add the identity matrix appropriately to the original equation). Well yes, however, the problem with square roots for matrices is that they are not unique (up to a plus or minus sign). So although the two unique scalar answers also work for the matrix equation [adding the appropriate identity matrices to the mix], the values are not unique.

To continue, just in case you are interested: When we solve the usual normalized scalar quadratic equation

. . .\(\displaystyle x^2\, +\, b\, x\, +\, c\, =\, 0\)

with the quadratic formula, we have really just completed the square to end up with an intermediate step

. . .\(\displaystyle \left[\, x\, +\, \left(\dfrac{b}{2}\right)\, \right]^2\, =\, \left(\dfrac{b}{2}\right)^2\, -\, c\)

If we take square roots of both sides and rearrange we have

. . .\(\displaystyle x\, =\, \dfrac{1}{2}\, \left[\,-b\, \pm\, \sqrt{\strut b^2\, - 4\, c\,}\,\right]\)

which is just the usual quadratic equation with a=1. Now we can reach basically the same point with matricies, i.e.

. . .\(\displaystyle X^2\, +\, b\, X\, +\, c\, I\, =\, 0\)

leads to

. . .\(\displaystyle \left[\, X\, +\, \left(\dfrac{b}{2}\right)\, I\,\right]^2\, =\, \left[\, \left(\dfrac{b}{2}\right)^2\, -\, c\, \right]\, I\)


[ X + (b/2) I ]² = [(b/2)² - c] I
This is where the difference between matrices and scalars start. Even if we assume the unique square root of A² is A (up to a sign) [it isn't necessarily, see the link above], we still have this leading to

. . .\(\displaystyle X\, =\, \pm\dfrac{1}{2}\, [\, -b\, I\, \pm\, \sqrt{\strut b^2\, - 4\, c\,}\, \sqrt{\strut I\,}\, ]\)

and the question of what is the square root of the matrix \(\displaystyle I\). Actually the square root of the 2X2 identity matrix \(\displaystyle I\) has an infinite number of solutions (see the link above)

. . .\(\displaystyle \sqrt{\strut I\,}\, =\, \begin{pmatrix} a&b\\c&-a\end{pmatrix};\, \mbox{ }\, a^2\, +\, b\, c\, =\, 1\)

besides the usual

. . .\(\displaystyle \sqrt{\strut I\,}\, =\, \begin{pmatrix}1&0\\0&\pm1\end{pmatrix}\)

and

. . .\(\displaystyle \sqrt{\strut I\,}\, =\, \begin{pmatrix} \pm1&0\\0&1\end{pmatrix}\)