Undergrad Factorization of a matrix equation

[Mr Davis 97]

This might be a dumb question, but I am wondering, given the equation $A\vec{x} - 7\vec{x} = \vec{0}$, the factorization $(A - 7I)\vec{x} = \vec{0}$ is correct rather than the factorization $(A - 7)\vec{x} = \vec{0}$. It seems that I can discribute just fine to get the equation we had before using the second $(A - 7)\vec{x} = \vec{0}$, so I'm not sure why I would think to do $(A - 7)\vec{x} = \vec{0}$ rather than $(A - 7I)\vec{x} = \vec{0}$.

 

[andrewkirk]

Yes you are right that that is the strictly correct way to write it. However the slight abuse of notation $(A-7)\vec x$ is sometimes used, because it is shorter to write and it is usually clear what it means. In that case the symbol 7 is interpreted to mean the operator on the vector space $V$ that maps $\vec v$ to $7\vec v$.

 

[Mr Davis 97]

Yes you are right that that is the strictly correct way to write it. However the slight abuse of notation $(A-7)\vec x$ is sometimes used, because it is shorter to write and it is usually clear what it means. In that case the symbol 7 is interpreted to mean the operator on the vector space $V$ that maps $\vec v$ to $7\vec v$.

Actually, a better question that I might ask would be that since in $A\vec{x} - 7\vec{x} = \vec{0}$ we have a matrix times and vector and then a scalar times a vector, what allows us to be able to factor out the vector? Wouldn't we get a matrix minus a scalar?

 

[andrewkirk]

No, because there is no rule that allows us to do that factorisation. It can only be factorised if we interpret the 7 as a linear operator, meaning it is $7I$.

 

[Mark44]

Actually, a better question that I might ask would be that since in $A\vec{x} - 7\vec{x} = \vec{0}$ we have a matrix times and vector and then a scalar times a vector, what allows us to be able to factor out the vector? Wouldn't we get a matrix minus a scalar?

Which is why you need to append I in the factorization.
In the expression $A\vec{x} - 7\vec{x}$ Ax is a vector and 7x is a vector, but if you factor the left side to (A - 7), then you're subtracting a scalar from a matrix. As you note, this doesn't make sense unless we stretch things to interpret 7 in the way that andrewkirk mentions. In this case 7 is really 7I.

 

[Mr Davis 97]

Why are we allowed to interpret 7 as either a scalar 7 or a matrix 7I? It seems somewhat ambiguous

 

[Mark44]

Why are we allowed to interpret 7 as either a scalar 7 or a matrix 7I? It seems somewhat ambiguous

As andrewkirk said in post #2, this is an abuse of notation, but when it is used, the context usually makes it clear what is intended.

However, every linear algebra book I've seen will write the factorization of Ax - 7x (for example) as (A - 7I)x, to show explicitly that we're not subtracting a scalar from a matrix.