linear quiz/ test 3

not true

LESS THAN OR EQUAL TO THE ALGEBRAIC MULT.

number of times an eigenvalue repeats. Or the number of times that a root appears in the characteristic polynomial.

dimension of eigenspace (also equal to number of free variables)

A square matrix A is said to be diagonalizable if A is similar to a diagonal matrix, that is, if A = PDP^-1 for some invertible matrix P and some diagonal matrix D

- a matrix is diagonalizable if the dimensions of its eigenvalues add up to the number of its columns AND rows??? (idk if its just columns and rows or both) see last question in 4.4 on page 4 titled the diagonalization theorem!!! for a great example

the eigenvalues of a triangular matrix are the numbers that go down diagonally

number of rows and columns that the corresponding matrix has and should also equal the degree of the matrix's characteristic polynomial

EQUAL TO 1

0 can NOT be on of its eigenvalues

0 CAN be one of its eigenvalues 4.3 pg 3

A scalar lambda such that Ax = lamba x has a solution for some nonzero vector x

a nonzero vector x such that Ax = lambda(x) for some scalar lambda

i.e. if you get 3 total eigenvectors for a matrix with 3 rows and 3 columns then that matrix is DIAGONALIZABLE!!!

(A hint for this is if a matrix has all different # eigenvalues, you can tell that it will have the same # of distinct eigenvalues which will add to the # of rows and columns naturally) 4.4 pg. 3

the D matrix just writes out the eigenvalues diagonally surrounded by all zeros.

read again 4.4 pg 4

DONT FORGET TO WRITE UR BASIS AS { } and ur EIGENSPACE AS =span ([ ]) !!!! READ CAREFULLY AND USE CORRECT NOTATION

to find the degree of a polynomial you add up the total of EVERY exponents (even the 1)

(1 + ¥)^3 (1-¥)^2 (¥ - 2) = 3 + 2 + 1 = 6 EXPONENTS

You'll know bc the characteristic polynomial you get will be impossible to factor (there's no values that multiply to 5 and add up to 4) and you'll have to resort to the quadratic formula

ex: (¥^2 - 4¥ + 5) is impossible to factor and doing the quadratic formula out gives its complex eigenvalues

the determinant of the TRANSPOSE and the determinant of the regular A matrix ARE THE SAME

if A is invertible and for example equal to A^-1, then ¥^-1 is automatically an eigenvalue for A^-1

Additional problems #2

Know that -Ais equal to regular matrix A (SEE PHOTOS)

vectors that equal zero when multiplied together (dot product is zero)

A list of vectors is called orthonormal if each vector is:

1) orthogonal (dot product equals out to zero)

AND 2) each vector has a length of 1

!!!! likely will be a MCQ (see pg. 3 of 5.2)

see pg 2 5.2

y= (y•u1/ u1•u1)(u1) + (y•u2/u2•u2) (u2) see pg 2 of 5.2

being asked to find a basis for the col (A __|__) AKA write the basis for the column space of the TRANSPOSE see pg 2 of 5.2

a basis for W that is also an orthogonal set ( vectors are proven to be an orthogonal set when their dot products equal out to zero, them equaling out to zero proves that the vectors are l.i. And in turn, when the # of vectors matches the dimension of the span i.e. 3 l.i. vectors in R^3, the vectors must be a basis for R^3 ) see pg 1 of 5.1

- a linearly dependent vector like the zero vector can't be among the vectors

- the vectors can't equal each other when multiplied together

Y= (y-^y) +y (use given y - found y hap, plus given y

IF YOU’RE ASKED TO WRITE THE INVERSE OF A MATRIX AND YOU ALR KNOW THAT IT IS ORTHOGINAL, THEN WRITE ITS TRANSPOSE AS THE ANSWER

if asked for basis for span R², you better write a basis with exactly TWO VECTORS (the basis is different than the span and doesn’t need a minimum of i.e. 2 vectors like the span does but instead needs exactly 2 vectors!!!)

basically saying that a vector with eigenvalue (1+2i) also has (1-2i) as an eigenvalue bc they are the SAME THING

the equation Ax=0 has only the trivial solution