Examples of using Lambda times in English and their translations into Polish
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It's equal to lambda times the vector.
Lambda times that non-zero vector v.
So if I take the determinate of lambda times the identity matrix minus A,
Lambda times the identity matrix times A.
So this is equivalent to the matrix lambda times the identity matrix minus A times the vector v.
Lambda times the identity matrix minus A ends up being this.
only if the 0 vector is equal to lambda times the identity matrix minus A times v.
It's lambda times the identity minus A.
the determinant of lambda times the identity matrix minus A is equal to 0.
This is lambda times the identity matrix in R3.
then this right here tells us that the determinant of lambda times the identity matrix,
Lambda times my eigenvector minus A times my eigenvector.
Or if we could rewrite this as saying lambda is an eigenvalue of A if and only if-- I will write it as if-- the determinant of lambda times the identity matrix minus A is equal to 0.
Lambda times this is just lambda times all of these terms.
And we used the fact that lambda is an eigenvalue of A, if and only if, the determinate of lambda times the identity matrix-- in this case it's a 2 by 2 identity matrix-- minus A is equal to 0.
And we get lambda times the identity matrix minus A times my eigenvector have got to be equal to 0.
then the determinant of lambda times the identity matrix minus A,
The null space of lambda times the identity matrix. And by an identity matrix minus A.
So this is true if and only if-- let's just subtract Av from both sides-- the 0 vector is equal to lambda- instead of writing lambda times v, I'm going to write lambda times the identity matrix times v.
So lambda times the identity matrix minus A is going to be equal to-- it's actually pretty straightforward to find.
So the eigenspace for lambda is equal to minus 1 is going to be the null space of lambda times our identity matrix,
Lambda times-- instead of v I will write the identity matrix, the n by
We said that if you were trying to solve A times some eigenvector is equal to lambda times that eigenvector, the two lambdas,
We know lambda times some eigenvector is the same thing as lambda times the identity matrix times that eigenvector.
So it's lambda times 1 is lambda, lambda times 0 is 0, lambda times 0 is 0, lambda times 1 is lambda. .
So that's the identity matrix times lambda.
Lambda squared times lambda is lambda cubed.
It's 0 times v1 plus lambda 2 times v2 and then plus 0 times everything else.
So it's going to be 4 times lambda minus 2 and we're subtracting.
Our characteristic polynomial has simplified to lambda minus 3 times lambda squared minus 9.