Theorem 4.2.1 · Standard matrix
If is linear, there is a unique matrix with for all , namely
Why: , so — a linear combination of the columns, which is exactly .
Theorem 4.2.3 · General bases
Let be an ordered basis for and for . For a linear there is a unique matrix with , and the th column of is .
★Key idea
Both bases matter, and they play different roles. decides what you feed in (apply to each ); decides how you record the output (express each result in -coordinates). Getting this backwards is the standard mistake.
Corollary 4.2.4 · The computational shortcut
For with bases and , the matrix representation is obtained by row reducing
It's the trick again — solving systems with the same coefficient matrix simultaneously.
Building a matrix representation
- 1Identify the two bases
for the domain, for the codomain. If none is stated, both are standard and you're in the easy case.
- 2Apply to each domain basis vector
Compute as ordinary vectors in .
- 3Express each result in -coordinatesF standardNothing to do — the vectors are their coordinates. Stack them as columns.F generalRow reduce to .
- 4Sanity check on one vector
Pick a , compute directly, and compare with re-expanded in . Two minutes here catches sign and ordering errors.
Worked example
0/5 stepsSame operator, non-standard bases
Let on . Find its matrix relative to in the domain and in the codomain.
∞Connection
If and you change basis with transition matrix , the operator's matrix changes from to — the definition of similar matrices. Theorem 6.1.1 then says similar matrices share a characteristic polynomial, which is exactly the statement that eigenvalues are a property of the operator, not of the coordinates.
Check yourself
Q1
To build the standard matrix of a linear transformation , the th column of is:
Q2
You need the matrix of relative to bases and . The efficient route is:
T / F
The matrix of the counterclockwise rotation of by is .
T / F
The same linear transformation can be represented by different matrices depending on the bases chosen.