Undergrad Can You Add a Scalar to a Matrix Directly?

[Danijel]

So, I recently came across this example: let us "define" a function as ƒ(x)=-x³-2x -3. If given a matrix A, compute ƒ(A). The soution proceedes in finding -A³-2A-3I where I is the multiplicative identity matrix.
Now , I understand that you can't add a scalar and a matrix, so the way I see it is that when saying A-3, we really mean A-3I, where the size of the matrix I is determined out of the context. But to me this is really non-intuitive. Actually, I see -3 as a separate function, say h(x)=-3, which is a matrix, so we actually have a constant matrix whose all elements are pairs((i,j),-3), or seeing it as a table, "all -3s". So, what is going on?

 

[scottdave]

I thought it was a little bit of hand waving the first time I saw that used in computing Eigenvalues.

I guess one thing you could do to think about it is first multiply both sides of ƒ(x)=-x³-2x -3 by an appropriate size Identity matrix.
Now you have Iƒ(x)=I(-x³)-I(2x) -I(3). Then go and use the A matrix instead of x, perhaps?

 

[fresh_42]

So, I recently came across this example: let us "define" a function as ƒ(x)=-x³-2x -3. If given a matrix A, compute ƒ(A). The soution proceedes in finding -A³-2A-3I where I is the multiplicative identity matrix.
Now , I understand that you can't add a scalar and a matrix,...

At least not in this context within a fixed vector space.

... so the way I see it is that when saying A-3, we really mean A-3I, where the size of the matrix I is determined out of the context.

Correct.

But to me this is really non-intuitive. Actually, I see -3 as a separate function, say h(x)=-3, which is a matrix, so we actually have a constant matrix whose all elements are pairs((i,j),3), or seeing it as a table, "all 3s". So, what is going on?

Square matrices of a certain size form a non commutative, associative algebra with $1$. As $1 \cdot A = A$ has therefore to hold, it is clear that we can identify $1=I$ with the identity matrix. One could also argue by the uniqueness of $1$. So if we identify all $\lambda \cdot 1 = \lambda \cdot I$ we have a natural embedding of the scalar field in this algebra. It makes sense and is usually assumed, because otherwise we have to be careful how to define scalar multiplication. I haven't checked, whether this identification is actually necessary to have all other axioms in place, or just a matter of convenience. You can try to find out by yourself. So $\lambda =\lambda \cdot I$ is simply a sloppy notation for the embedding. The all $3$ matrix in your example would contradict the algebra axioms (I guess) for the one-element $I$.

That's the true reason modulo some considerations about necessities. For (very) short: $\lambda =\lambda \cdot I$ is simply a sloppy notation.